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intend this book to be sold to the Public 
at the advertised price, and supply it to 
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of discount. 



ELECTRICAL 



AND 



MAGNETIC CALCULATIONS 



FOR THE USE OF 



Electrical Engineers and Artisans, Teachers, Students, 
and all others interested in the Theory and Ap- 
plication of Electricity and Magnetism 

A?AKATKINSON, M.S. 

Professor of Physics and Electrical Engineering in 
Ohio University, Athens, Ohio. 



FOURTH EDITION, REVISED 




* * 

• • • 



NEW YORK: 

D. VAN NOSTRAND COMPANY 
25 Park Place 

1913 



QCS32 



COPYRIGHT „ ig02. IQ03, I913 
BY 

D. Van Nostrand Company 



C. J. PETERS & SON, TYPOGRAPHERS 
BOSTON, MASS., U. S. A. 



CI.A346055 

U-01 



TABLE OF CONTENTS. 



CHAPTER I. page 

Explanation of Units I 

CHAPTER II. 
Relation of Quantities 12 

CHAPTER III. 
General Laws of Resistance 22 

CHAPTER IV. 
Electrical Energy 45 

CHAPTER V. 
Wiring for Light and Power 66 

CHAPTER VI. 
Batteries 85 

CHAPTER VII. 
Magnetism ...,,-.... no 



iv CONTENTS 

CHAPTER VIII. 
Relation of Magnetic Quantities 116 

CHAPTER IX. 
The E.M.F. of Dynamos and Motors 148 

CHAPTER X. 
Calculation of Fields 171 

CHAPTER XL 
Elements of Dynamo Design 190 

CHAPTER XII. 
Alternating Currents 230 

CHAPTER XIII. 
Alternating Current Distribution 261 



PREFACE. 



The following pages, both in plan and material, are the 
outgrowth of several years of experience in teaching young 
men the rudiments of electricity. A large part of the 
matter, in fact, was prepared expressly as an introduction 
to a course in electrical engineering ; since there was noth- 
ing published covering the topics found desirable, and 
making use of the method herein employed. 

A multiplicity of wordy rules and unexplained con- 
stants arbitrarily set down burden the memory unneces- 
sarily, are often unintelligible to the reader, and are at 
best very clumsy tools with which to work. In the pres- 
ent volume, on the other hand, several processes are 
brought together wherever possible under a single broad 
principle, which is then expressed by means of a formula. 
The treatment in this respect aims to be educational. 
Through a step by step process principles and formulae 
are evolved from facts and principles already understood. 
After the law has been clearly developed, and has been 
given the most concise, easily remembered, and con- 
venient working form, the method of induction gives way 
to that of deduction. A series of examples are then 
worked out illustrating the application of the principle, 
and giving familiarity with its processes. At the end of 
the chapters are also lists of original problems for drill 
in the mastery of the principles and their application. 



VI PREFACE. 

It is hoped that the great body of artisans in all the 
departments of electrical engineering practice will find in 
these pages an invaluable aid in their efforts to acquire 
a better working knowledge of the principles underlying 
their profession. In a first reading, perhaps the original 
problems may be passed over ; also the chapter on the 
relation of heat and chemical energy, and possibly a 
portion of alternating currents, especially that of long- 
distance transmission circuits. Each individual's taste 
and particular line of work will largely guide him in his 
selection of subjects for special investigation. A careful 
study of this little volume, even omitting certain portions 
as suggested, will be an excellent schooling for those 
busily engaged in the various electrical processes. 

It will also be a helpful companion to electrical engi- 
neers, superintendents, and all those in the more respon- 
sible positions in engineering work. They will find it 
valuable in its development of the rules and formulae 
employed in their profession and as a handy reference 
for the methods of application of these rules and formulae 
to practical engineering problems. 

Teachers in schools and colleges which devote some 
time to the subject of physics may find here a vast num- 
ber of examples for class use in teaching electricity, the 
most important branch of physics. It will be particularly 
useful to teachers in colleges and technical schools which 
make a special feature of electricity, either as a reference 
book of formulae and examples, or as a text-book for class 
drill in those topics treated. Selections may be made of 
the topics best suited to particular needs, if not all are 
available. 



PREFACE. Vll 

In the preparation of this book frequent use has been 
made of competent authorities. Most of these have been 
mentioned in the body of the work, or in the marginal 
references. 

Mistakes in a book of this nature are inevitable, how- 
ever carefully the proof has been read, and the author will 
appreciate any corrections sent to him. 

This opportunity is taken of thanking our friend and 
professor, W. M. Stine, Professor of Engineering in 
Swarthmore College, for examination of the manuscript 
and valuable suggestions in connection with its publi- 
cation. Acknowledgments are also due to Professor 
W. B. Bentley, Department of Chemistry, Ohio Univer- 
sity, for assistance in reading the proof, and to Messrs. 
F. H. Super and N. R. Cunius, Assistants in the Depart- 
ment of Physics and Electricity, Ohio University, for the 
preparation of the diagrams. 

Ohio University, 
Jan. /, 1902. 



PREFACE TO THE SECOND EDITION. 



The author has been gratified by the many appreci- 
ative reviews and notices published in reference to the 
first edition of this volume. It is not a treatise on Elec- 
trical Engineering, but, as its title indicates, attempts to 
set forth the methods of solution of those problems in 
Electricity and Magnetism which are of most importance 
to the electrical engineer and the teacher. As such it 
has been most kindly received. 

Coming so soon after the publication of the first edi- 
tion perhaps not all the errors have been discovered and 
corrected, and but few changes are considered necessary, 
otherwise. 

The chapter on Alternating Currents has been moved 
to the end of the book, and Chapter XIII., Alternating 
Current Distribution, formed from the portion of Chapter 
VI., relating to Alternating Currents, added. These modi- 
fications will no doubt improve the usefulness of the 
book. 

Ohio University, 
Oct. i, 1902. 



PREFACE TO THE FOURTH EDITION 



The present edition of this text has been improved in 
the following particulars : A considerable number of errors 
in the former edition have been noted and corrected in 
this ; some terms made obsolete by common usage of a 
substitute, such as " duty " and " weber," have been 
changed to the modern terms ; in Chapter V constants 
are suggested by whose use the rules for wiring, applicable 
to carbon lamps, may also be adapted to tungsten lamps 
now in common use; also it is suggested on pages 143, 
147 and 199 that the original problems in Chapters VIII, 
X and XI, in which magnetomotive force or ampere-turns 
are required, be worked out by the use of the curves on 
page 229, or by the use of the tables given on page 298. 
From the curves the value of H corresponding to any 
value of B is obtained for each portion of the magnetic 
circuit ; each H (gilberts per centimeter) is then multi- 
plied by the corresponding length in centimeters, giving 
the magnetomotive force, whence the sum of all the 
partial magnetomotive forces multiplied by eight-tenths 
will give the total ampere-turns required ; that is, ampere- 
turns = 0.8 M. From the tables on page 298 the " am- 
pere-turns per centimeter'' are obtained direct - — 

whence the ampere-turns for each portion and the total 
for the magnetic circuit may readily be computed. These 



X PREFACE. 

methods will give results only approximately the same 
as those obtained by the more cumbersome reluctance 
method, for reasons which will at once be apparent. 
Their convenience and practicability recommend their 
general use for all such computations. 

It is to be hoped that those accustomed to the use of 
this book, as well as others who may use it, will find it 
more satisfactory because of the changes and suggestions 
made in this edition. 

Ohio University, Athens, Ohio, 
Jan. 2, 1913. 



ELECTRICAL AND MAGNETIC 
CALCULATIONS. 



I. 

EXPLANATION OF UNITS. 

i. Units. — In every form -of measurement certain 
units are necessary as standards of comparison. A quan- 
tity of any kind is measured when the number of the 
units it contains is determined. Thus a bin of wheat 
is measured when the number of bushels it contains is 
found, the bushel being the unit. The quantity of elec- 
trical flow is measured by expressing the number of 
coulombs in it. The expression of a quantity, therefore, 
includes the name of the unit employed, preceded by the 
number of units ; as 40 volts, or 5 amperes. 

2. Fundamental and Derived Units. — The fundamental 
units of any system are those which are basal, not derived 
from any other units. There are considered generally 
three fundamental units, — the unit of time, the unit of 
mass, and the unit of length. In one system these are 
respectively, — the second, the gram, and the centi- 
meter. These are the units used for all scientific work. 
Other units built up by combining these units are called 
derived units. Such are units of area involving the sec- 



2 ELECTRICAL AND MAGNETIC CALCULATIONS. 

ond power of a length, P ; also units of volume being pro- 
portional to / 3 . Likewise the watt is a derived unit, being 
a product; its factors are an electromotive force and a 
current, or E X /, each of which in turn is derived. 

3. Absolute Units. — If the fundamental units men- 
tioned above be used, the system of absolute units based 
upon them is called the C.G.S. system, or the centimeter- 
gram-second system, from the names of the fundamental 
units of the system. Thus the dyne is the absolute unit 
of force, being that force which, acting upon one gram 
for one second, will give it a velocity of one centimeter 
per second. The erg is the absolute unit of work. It is 
the w r ork done when one dyne acts through a distance of 
one centimeter. 

It is to be observed that a unit is denned by unit con- 
ditions ; that is, by the units of the elements upon which 
the unit under definition is based. To illustrate, work 
involves two, and only two elements ; namely, force and 
distance. Hence, unit work will be defined by unity in 
these elements. The erg is the work of one dyne through 
one centimeter, 

4. The Basis of the Fundamental Units. — {a) Origi- 
nally the meter was intended to be the ten-millionth part 
of the earth-quadrant through the meridian of Paris, 
measured from the equator to the north pole. For practi- 
cal purposes, the French government adopted a platinum 
standard between whose ends at o° C. the distance should 
be one meter. This is known as the " Metre des 
Archives." It was made by Borda in accordance with a 
government decree passed in 1795. In 1866, the United 



EXPLANATION OF UNITS. 3 

States government, by Act of Congress, defined the meter 
to be 39.37 inches. The ce?itimeter is the one-hundredth 
part of the meter ; the millimeter is the tenth part of the 
centimeter, or the one-thousandth part of the meter. For 
larger measures the kilometer is employed, and is equal to 
1000 meters. 

(b) Mass is the quantity of matter in a body, while its 
weight is the amount of the earth's force of gravity upon 
the mass. The mass does not vary, but weight varies 
from place to place due to changes in the force of gravity. 
Calling the force of gravity g, representing mass by m, 
and weight by w, this relation is expressed by 

w = mg. (1) 

Weight is therefore a function both of mass and the force 
of gravity. In the C.G.S. system the unit of mass is 
theoretically that of a cubic centimeter of distilled water 
at the temperature of its greatest density, 4 C, and is 
called the gram. As a practical standard, however, it is 
the one-thousandth part of a mass of platinum preserved 
in the archives of Paris, called the " Kilogram des 
Archives." The reason for using the larger material 
standards for length and mass is obvious. The kilogram 
has the value in English measure of 2.20462 pounds, or 
approximately 2\ lbs. 

(c) The second is the -^J^o P a]rt °f tne average length 
of all the solar days taken throughout the year. A solar 
day is the interval between two successive transits of the 
sun's center across the meridian of a place. These days 
vary in length on account of the unequal velocities of the 



4 ELECTRICAL AND MAGNETIC CALCULATIONS. 

earth in its orbit from day to day. Hence the average of 
all is taken throughout the year. 

5. Magnetic Units. — ( a ) The Strength of Pole of a 
magnet is determined by the force it is capable of exerting 
on another pole at a given distance. If m be the strength 
of one pole and tn r that of another at a distance of d, then 
the force of attraction or repulsion, as the case may be, 
is expressed by the formula, 

mm' 

in which q is a constant. The force, therefore, varies 
directly as the strength of the poles and inversely as the 
square of their distance apart. A unit pole is one such as 
to repel a similar one at a distance of one centimeter with 
a force of one dyne. 

(b) Magnetic Field. Any space traversed by magnetic 
forces is called a magnetic field. The space between the 
poles of a horse-shoe or other magnet is conceived to 
be traversed by magnetic lines of force ; that is, imaginary 
lines along which magnetic attraction or repulsion takes 
place. If the force is one dyne per square centimeter of 
the surface normal to the direction of the lines of force, 
we say the field has an intensity of one ; if the force be 
10 dynes per square centimeter, the intensity is 10, or 10 
lines of force per square centimeter, etc. Intensity of 
field is represented by H. If a pole of strength m be 
put into a field whose intensity is H, the force tending to 
push it along the lines is 

F=Hm. (3) 



EXPLANATION OF UNITS. 5 

(c) Magnetic Moment. The magnetic moment of a 

magnet is the product of its strength of pole by its 

length, or 

M = ml. (4) 

Suppose a slim magnet of length / be placed at right 
angles to the lines of force of a field of intensity H, then 
the moment of the couple tending to swing the magnet 
around into parallelism with the lines of force will be 

F = Htnl = HM. (5) 

(d) Intensity of Magnetization. The quotient obtained 
by dividing the magnetic moment by the volume of the 
magnet is called the intensity of magnetization. 

_ M ml m , N 

i= v = ti=i- (6) 

In other words, it may be defined as the strength of pole 
per unit surface of cross section. 

(e) Magnetic Induction. When a piece of soft iron is 
placed parallel with the lines of force of a magnetic field, 
the lines concentrate within the bar, which becomes a 
magnet by i?iduction. The end at which the lines enter be- 
comes a south pole, the end from which they leave a north 
pole. The iron offers a path of less resistance to the mag- 
netic lines than air does. This is expressed by saying that 
iron has a smaller reluctivity, or a greater permeability, 
than air. When speaking of complete magnetic circuits 
the terms reluctance and permeance are used instead of the 
above, which apply to specific substances. Air is taken 
as the standard of permeabilities, its own being therefore 



6 ELECTRICAL AND MAGNETIC CALCULATIONS. 

unity. The number of lines of force passing through a 
given length of different substances is. proportional to 
their respective permeabilities, the areas of cross section 
being i square centimeter. If H represents the intensity 
of the lines in air, and the intensity of induction in a 
piece of iron put into the field H be expressed by B, then 
under the conditions the permeability of the iron will be 

B_ 

from which B = fxH. (7) 

If the bar, say, steel, be already magnetized to intensity 
7, when placed in the field Zf, the induction will become 

B= IT+qvI. (8) 

6. Electrical Units. — There are two systems of electri- 
cal units : (1) units founded on the force of attraction of 
unlike or repulsion of like charges of electricity, constitut- 
ing the electrostatic system ; (2) units based on considera- 
tions of the magnetic field produced by a current of 
electricity flowing in a wire, constituting the electromagnetic 
system. All our practical electrical units are derived 
from the latter system, which is, therefore, the only one 
that will be considered. 

( a ) The Unit of Current. The absolute unit of current 
is defined as that rate of flow in a conductor bent into the 
form of a ring whose radius is one centimeter, which will 
exert a force of one dyne at the center for each centimeter of 
the arc. Hence, the radius being unity, the whole cir- 
cumference, 27rr, will be 6.28 cm., and the total force 



EXPLANATION OF UNITS. 7 

exerted at the center 6.28 dynes. This unit is the abso- 
lute ampere. Suppose, now, a loop of any radius r be 
made to carry any number of absolute amperes /; the 
force at the center of the loop in dynes will be 

2 irrl 2 ttI I 

^=-72-=— = 6 - 28 "- (9) 

If the circular coil have n turns instead of one, the force 

becomes 

2iznl 
F=——- (10) 



In case the wire is wound in a long solenoid whose length 
/ is great compared with its radius, the force inside it 
will be 

F =— r— ( IJ ) 



This unit is large, and for practical purposes the ampere 
is chosen equal to -^ of the above unit, so that if one 
practical ampere flow in the loop as described the force 
at the center due to the whole circumference will be 
6.28 X io _1 = 0.628 dyne. 

The Chamber of Delegates of the International Con- 
gress of Electricians which convened in Chicago, Aug. 
21, 1893 (World's Fair), defined the unit of current 
as follows : 

As the unit of current, the international ampere is recommended 
to be adopted which is one-tenth of the unit of current of the C.G.S. 
system of electromagnetic units, and which is represented sufficiently 



8 ELECTRICAL AND MAGNETIC CALCULATIONS. 

well for practical use by the unvarying current which, when passed 
through a solution of nitrate of silver in water, in accordance with 
accompanying specifications, deposits silver at the rate of 0.001118 
gram per second. 

If the solution be copper sulphate the international 
ampere will deposit copper at the rate of 0.0003284 gram 
per second under specific conditions. These constants, 
0.001118 and 0.0003284, are called the electrochemical 
equivalents of silver and copper, respectively. Sub-divis- 
ions of the ampere are often used for the measurement of 
very small currents ; as the milliampere, or the T qW of 
the ampere, and the microampere, or the y^olxffoTr °f tne 
ampere. 

(b) The Unit of Quantity. The quantity of electricity 
conveyed is expressed by the product of the current and 
the time flowing in seconds. Unit quantity will be trans- 
ferred when one ampere flows for one second. This is called 

the coulomb and is equal to io _1 x 1, or io _1 C.G.S. units. 
Suppose 5 amperes flow for 4 seconds ; £=5x4 = 20 
coulombs, or 20 X io _1 = 2 C.G.S. units of quantity. 
Microcoulombs are used to express small quantities. 

(c) The Unit of Electromotive Force. The general defini- 
tion of force is that which tends to produce or modify 
motion. By analogy, one would say that electromotive 
force is that which tends to move or transfer a quantity of 
electricity. It is analogous to pressure produced by a 
head of water, as in a standpipe. If a tap be opened 
somewhere in the street water-main there will be a flow 
through the pipe because the pressure at the tap is less 
than that at the standpipe. Let E x be the pressure at the 



EXPLANATION OF UNITS. 9 

latter, and E 2 the pressure at the tap. The quantity flow- 
ing out is proportional to E 1 — E 2 = E. This difference of 
pressure, which may be represented simply by E, may be 
called the watermotive force. Similar conditions exist in 
the phenomenon of electromotive force, or electric pres- 
sure. All that the battery or dynamo does is to set up a 
difference of potential, or difference of electric pressure 
between two points, called electromotive force E. The 
phenomenon of the equalization of this difference in 
a circuit is what we mean by current flow. Now, if 
the opposing forces, or resistance, be unity, the unit of 
E.M.F. is that which will produce unit flow of current 
through the circuit. 

The absolute unit of force being very small, the practical 
unit of E.M.F. is chosen equal to 100,000,000 or io 8 C.G.S. 
units, and is called the volt. The International Congress, 
mentioned before, recommended for adoption, 

As the unit of electromotive force, the international volt, which is 
the E.M.F. that steadily applied to a conductor whose resistance is 
one inter7iational ohm, will produce a current of one international 
ampere, and which is represented sufficiently well for practical use by 
\%% 4 °f tne E.M.F. between the poles of the voltaic cell, known as 
Clark's Cell, at a temperature of 15 C, and prepared according to 
specification. 

The Clark cell has therefore 1.434 international volts of 
E.M.F. The Carhart cell, which is much superior to the 
old Clark, is now much used in this country as a stan- 
dard. It has an E.M.F. of 1.44 volts at 15 C. 

One volt will be generated in a conductor which is 
moved across a magnetic field so as to cut the lines of 
force at the rate of io 8 per second. 



10 ELECTRICAL AND MAGNETIC CALCULATIONS. 

( d ) The Unit of Resistance. Resistance to the flow of a 
current is of the nature of an opposing force. The unit 
of resistance is such that one volt of E.M.F. will cause 
current to flow through it at the rate of one ampere 
per second. The practical unit is chosen equal to io 9 
C.G.S. units of force and is called the ohm. The adopted 
international ohm 

Is based upon the ohm equal to io 9 units of resistance of the C.G.S. 
system of electromagnetic units, and is represented sufficiently well 
for practical use by the resistance offered to an unvarying electric 
current by a column of mercury at the temperature of melting ice, 
14.4521 grams in mass, of a constant cross sectional area, and of the 
length 106.3 centimeters. 

For practical standards, coils of constantan or manga- 
nin wire are wound on spools, standardized and placed in 
a suitable box, the coils being connected to brass bars on 
the top, so that by using plugs any resistance within the 
capacity of the box can be obtained. The megohm, or 
1,000,000 ohms, is used in the measurement of high resis- 
tances, such as the resistance of insulators. For very 
small resistances the microhm, or T o onm * s used. 

( e ) The Unit of Capacity. If one coulomb of electricity 
be stored in a recipient, for instance, in a coil of insulated 
wire, or in a system of flat parallel conductors, adjacent 
ones insulated from each other, called a condenser, and if 
this quantity tends to escape with an E.M.F. of one volt, 
the capacity of the recipient or condenser is unity. This 
unit is the farad. Expressed in another way, it is the 
capacity such that one volt will store in it one coulomb of 
electricity. For all ordinary capacities the microfarad, or 



EXPLANATION OF UNITS. II 

T<nr<io oo °f tne ^ ara( i is used, the farad being very large, 
io" 9 C.G.S. units. 

The capacity of the earth is T oo§ooo farad, 'or 636 
microfarads. A Leyden jar with a total tinfoil surface of 
1 square meter, and glass 1 millimeter in thickness, has a 
capacity of F V microfarad. 

(f ) The Unit of Power. The unit of electrical power is 
called the watt, and is the energy required to move one 
ampere per second through one ohm resistance. In other 
words, one volt of E.M.F. delivering one ampere of current 
per second represents a power of one watt. The watt is 
equal to io 7 ergs per second of C.G.S. units. The horse- 
power is 33,000 foot-pounds of mechanical work per minute. 
The watt is T ^g of the horse-power, and 1 H.P. = 746 
watts. 

(g) The Unit of Work. The unit of work is done when 
one watt of energy is expended per second, and is called 
the joule, which is also equal to 10 7 ergs. Joules of work 
are obtained by multiplying watts of power, or energy, by 
seconds of time. 

(h) The Unit of Induction. The induction is unity 
when the E.M.F. induced is one international volt, 
while the inducing current varies at the rate of one inter- 
national ampere per second. This unit is called the 
henry, and is io 9 C.G.S units. 

These international electrical units were legalized by 
Act of Congress, approved by the President, July 12, 1894, 
so that they take their place with other standards of 
weights and measures. 



12 ELECTRICAL AND MAGNETIC CALCULATIONS. 



II 



RELATION OF QUANTITIES. 

7. Ohm's Law. — The simplest perhaps, and yet the 
most important relation of electrical or magnetic quanti- 
ties to each other is that expressed in Ohm's law. This 
law expresses the relation of E.M.F., current, and resist- 
ance. Putting / for current, E for E.M.F., and R for 
resistance, the law* is expressed by the formula, 

Hence, current is obtained by dividing electromotive force 
by resistance. From the above formula, by simple trans- 
position, or by the rules of simple division, we obtain 

■p 
E = 7R, and R = — • In words, these mean that the 

E.M.F. is equal to the product of current and resistance, and 
that resistance is equal to the quotient of E.M.F. by current. 

Example. — How much current in amperes will flow 
through an incandescent lamp whose resistance is 200 
ohms, when an E.M.F. of no volts is applied to it ? 

Solution. — /= — = = 0.55 ampere. 

R 200 

Example. — A battery has a resistance r of 3 ohms ; 
what is its E.M.F. if it causes a current of 0.05 amperes 
to flow through an external resistance equal to 60 ohms ? 



RELATION OF QUAWTITIES. 1 3 

Solution. — E = / (R + r) = 0.05 (60 + 3) = 3.15 
volts. # 

Example. — Find the resistance of an arc street lamp 
when a voltage of 50 is required to send 7 amperes 
through it. 

Solution. — R = — = ^— = 7 - ohms. 

/ 7 7 

Expressed in C.G.S. units, ^ = io 8 , R = io 9 , and 

C 8 

1 =z —= — - = io -1 or yL. That is, the ampere is the 

r \j- of the C.G.S. electromagnetic unit of current. 

8. Quantity, Electromotive Force and Capacity. — 

Example. — How many coulombs of electricity will flow 
through an incandescent lamp whose resistance is 200 
ohms, when an E.M.F. of no volts is applied to it for 

2 seconds? 

Solution. — Q=Ixt=-= X / = X 2 = 1.1 coulombs. 

R 200 

Example. — A glass plate has a square of tinfoil pasted 

on each side, thus forming a storehouse or condenser. 

A battery whose E.M.F. is 10 volts is connected to both 

sides by wires, and thus gives it a charge of 0.00000000 1 

coulomb = 0.00 1 microcoulomb : what is the capacity of 

the receptacle ? 

Solution. — From obvious considerations the quantity 
stored will depend on the pressure applied and the capa- 
city of the storehouse : that is 

Q=£C. (13) 

From this 

_ Q 0.000000001 

C = -=, = = 0.0000000001 farad 

E 10 

= 0.000 1 microfarad. 



14 ELECTRICAL AND MAGNETIC CALCULATIONS. 

Example. — How many volts will be required to charge 
a condenser whose capacity is 0.5 microfarad with 2.5 
microcoulombs of electricity ? 

Solution. — From equation (13) E = ~> Hence, in 

2 c 

this case, E = — = 5 volts, the required E.M.F. Or, 
since 2.5 microcoulombs = 0.0000025 coulomb, and 0.5 

r i r i ^ 0.000002 C 

microfarad =0.00000015 farad, E = - = c volts, 

0.0000005 

as before. 

Example. — Find the quantity in microcoulombs which 
a cell of 1.8 volts E.M.F. will store in a condenser whose 
capacity is 0.1 microfarad. 

Solution. — Q —EC = 1.8 X 0.1 = 0.18 microcoulomb. 

Example. — Find the value of the last answer in abso- 
lute C.G.S. units. 

Solution. — The coulomb = io _1 C.G.S. units of 
quantity. 0.18 microcoulomb = 0.00000018 coulombs = 
0.00000018 X io _1 C.G.S. units. 

9. Power and Work. — Example. — A storage battery 
has an E.M.F. of 26 volts; assuming the internal resist- 
ance r to be 2 ohms, and the external resistance R to 
be 50 ohms, how many joules of work will it do in 30 
minutes ? 

r E 2 6 

Solution. — 30 mm. = 1800 sec. I = — = 



R 2 + 50 

\ ampere. Watts = E X / = 26 x^ = i3. Joules 
watts X seconds = 13 X 1800 = 23,400 joules. 



RELATION OF QUANTITIES. I 5 

Or, Watts = — = — = 1 x ; and 

R + r 52 °' 

1800 x 13 = 23400 joules. 

Example. — If the indicator shows that an engine is 
developing 30 H.P., what is the output of the dynamo to 
which it is attached, neglecting all losses in both engine 
and dynamo ? 

Solution. — 30 H.P. = 30 x 746 = 22,380 watts = 2 2.38 
kilowatts, or K.W. 

Example. — Suppose the above dynamo supplies 400 
lamps at no volts: how much current must each lamp 
require, and how much work is done in each per night of 
10 hours ? 

Solution. — 22,380 watts -5- no volts = 203.45 am- 
peres. This amount is for 400 lamps. Hence each will 
receive 203.45 -s- 400 = 0.508 ampere. no volts X 
0.508 = 55.88 watts per lamp. 10 hours = 10 X 60 X 
60 = 36,000 seconds. Hence the total work done = 55.88 
X 36,000 = 2,011,320 joules per lamp. 

10. The Magnetic Relations of Current. — Example. — 
Let a circular loop of wire having a radius r of 5 cm. 
carry 10 C.G.S. amperes of current; how much force in 
dynes will be exerted at its center ? 

Solution. — F— = 6.28 -= 6.28 x — 

r r 5 

= 12.56 dynes. See equation (9). 

Example. — How many international amperes will be 



\6 ELECTRICAL AND MAGNETIC CALCULATIONS. 

required in a circular coil of 20 turns, radius of coil 
10 cm., to produce a force of 125.6 dynes at its center? 

Solution. — From (10), F = 6.28 — and / = 



r 6.28 n 

Therefore /= *° * 25 ' = 10 C.G.S. units. But the 
6.28 X 20 

international ampere is io" 1 C.G.S. units. Therefore 

10 C.G.S. amperes = 10 -5- io _1 =100 amperes. 

Example. — How many turns of wire will be necessary, 
and on what radius wound, to produce a force of 125.6 
dynes at the center when carrying 1 ampere ? 

Solution. — F= = 12 c. 6. 

lor D 

n 12C.6 x 10 i2c6 

Hence - = — - — = - — ^ = 200. 

r 2 -kI 6.28 X 1 

An indefinite number of answers is obviously possible. 
Within reason, however, a comparatively small number of 
solutions is permissible. For instance, make n = 400 ; 
then r = 2 cm. Or say n = 800, whence r = 4 cm. 
Again, n may be 1000 and r = 5.0 cm. 

Example. — How many dynes of force will be exerted 
inside a solenoid, or long coil, whose length is 20 cm., 
and the number of whose turns is 100, when it carries 10 
amperes of current ? 

Solution. — From (11), 
7? = 4?r = (12.56 X 100 X — J -4- 20 = 62.8 dynes. 

11. Electrolytic Effects. — Example. — How many am- 
peres of current have been flowing for 30 minutes when 



RELATION OF QUANTITIES. 1 7 

the amount of silver deposited by it upon the negative 
plate, or cathode, is found to be 0.0040248 gram ? 

Solution. — The current multiplied by the amount 
that 1 ampere in 1 second deposits, and this product by 
the time in seconds, gives the total weight deposited. 
Briefly, 

W=Izt, (14) 

in which W is the weight in grams deposited on the 
cathode ; z is the weight deposited per second by 1 am- 
pere ; and /is the time in seconds that the current is 
passing through the solution of silver nitrate, AgN0 3 . 
In this problem, / = 30 X 60 = 1800 seconds; z = 
0.001118 gram per ampere per second, or the electrochem- 
ical equivalent of silver; w = 0.0040248 gram. Hence 

,. W 0.0040248 

1 = — = — = 0.002 amp. 

zt 0.001118 X 1800 

Example. — How long must 1 ampere flow through a 
solution of copper sulphate, CuS0 4 , with copper elec- 
trodes, cathode, and anode, to deposit 0.3284 gram of 
copper ? 

Solution. — From (14), 

W 0.3284 „ . 

/ = —=- = — = 1000 sec. = 16 mm. 40 sec. 

Iz 1 X 0.0003284 

The electrochemical equivalent of copper is 0.0003284 

gram per ampere per second. 

Example. — How much water, H 2 0, will be decom- 
posed, that is, separated into its elements of O and H, by 
5 amperes flowing between platinum electrodes for 60 
minutes ? 



1 8 ELECTRICAL AND MAGNETIC CALCULATIONS. 

Solution. — The electrochemical equivalent of H = 
0.00001038 gram; of 0=0.00008283 gram. Hence 
1 ampere will decompose in 1 second, 0.00001038 + 
0.00008283 = 0.00009321 gram of water. Therefore, 
(14), W= Izt = 5 X 0.00009321 X 3600 = 1.67778 grams 
= 1.67 cubic centimeters at 4 C. 

12. Original Problems. — 1. What must be the external 
resistance of a circuit so that a battery of 1.8 volts 
E.M.F. and J ohm internal resistance will send 1.5 am- 
peres through it ? 

jR = 1 ohm. 

2. An incandescent lamp has a resistance, when hot, 
of about 220 ohms, and requires 1 ampere of current. 
If the dynamo has a resistance of -^ ohm, how many 
ohms may the lead wires have if the voltage of the 
dynamo is 111 ? ■ 

R = 1.9 ohms. 

3. What must be the E.M.F. of a generator to supply 
40 lamps each requiring ^ ampere, the resistance of the 
generator being 0.1 ohm, so that each lamp, resistance 
220 ohms, shall get its full current, assuming the wires to 
have a resistance of 0.5 ohm ? 

E = 122 volts. 

4. A storage battery consists of 10 cells in series, each 
giving 2 volts E.M.F. and having \ ohm internal resis- 
tance ; the external circuit consists of two resistance 
coils in series, one of 5^ ohms, the other 8 ohms ; how 
many amperes of current will flow in the circuit ? 

/= 1.25 amperes. 



RELATION OF QUANTITIES. 1 9 

5. When an E.M.F. of no volts is applied to a bank 
of 55 incandescent lamps in parallel, 27.5 amperes flow 
in the circuit ; how many ohms resistance in the bank ? 
How does this compare with the resistance of a single 
lamp ? What does this last illustrate ? 

R = 4 ohms. 

One lamp has ^f& =55 times 
as much as 55 lamps. 

6. How many coulombs would be used in the bank, in 
a 6 hours' run, and how much must be charged per 
ampere-hour in order to make the income equivalent to 
40/ per lamp per month, the month consisting of 30 
days, and 6 hours' run each day ? 

Q == 594,000 coulombs. 

Price = 0.44/ per ampere-houi. 

7. A cell whose E.M.F. is. 1 volt is used to charge a 
condenser which is afterwards discharged through a 
circuit in which is a milliamperemeter. The discharge 
required 1 second and the milliammeter read y 1 ^ milliam- 
pere. Find the capacity of the condenser in microfarads. 

C= 100 mf. 

8. What must be the E.M.F. of the charging battery 
when it is required that a condenser whose capacity is 
^ microfarad, on discharge gives 0.000002 coulomb 
through the ballistic galvanometer ? 

E = 6 volts. 

9. There are 500 lamps, no volts, 220 ohms each in a 
certain building, and they burn 5 hours out of each 24. 
How much must the company furnishing the power 



20 ELECTRICAL AND MAGNETIC CALCULATIONS. 

charge per watt-hour, meter rates, in order that it may 
realize the equivalent of 50/ per lamp per month, flat 
rate, assuming 30 days per month ? 

Price 0.0061/ per watt-hour. 

10. A dynamo generates 116 volts, of which no volts 
are required in the lamps, the other 6 being necessary 
to pass the current through the machine and line. The 
lamps have a resistance each of 220 ohms, and there are 
40 of them in parallel. What per cent of the watts is 
spent in the lamps, and what per cent as waste in the 
machine and line ? 

Per cent in lamps = 94.8. 
Per cent in waste = 5.2. 

11. Find the H.P. of the machine in problem 10, and 
the H.P. used in the lamps. 

H.P. of machine = 3.1. 
H.P. of lamps = 3 — . 

12. Reduce the machine energy in 10 to C.G.S. values 
and to joules of electrical work. 

Joules per second = 2320. 
Ergs per second = 2320 X io 7 . 

13. How many turns of wire will be necessary to make 
a solenoid 2 cm. diameter and 50 cm. long so that 
5 amperes will produce a field within which the force 
shall be 62.8 dynes ? 

n = 500 turns. 

14. Determine the amount of current that must flow in 
a circuit in which is a silver voltameter so that the 



RELATION OF QUANTITIES. 21 

cathode which weighs before passing the current 26.7571 

grams shall weigh, after passing the current 1 hour, 

26.9128. 

/= 0.03868 ampere. 

15. The above current passed through a tangent 
galvanometer whose deflection was 53.5° = 6. The law 
of the tangent galvanometer is / = K tan 6, where / is 
the current, K is the galvanometer constant, or the 
current necessary to cause a deflection whose tangent is 1 . 
Find K. 

K = 0.0286 ampere. 

16. Find the constant K of a small portable galvanom- 
eter in series with a copper voltameter, when the fol- 
lowing observations were taken : weight of cathode before 
test, 4.82465 grams ; after passing current 30 minutes, 
weight was 4.831 grams ; deflection, 6 = 53 . 

K = 0.0081 1. 



22 ELECTRICAL AND MAGNETIC CALCULATIONS. 



III. 

GENERAL LAWS OF RESISTANCE. 

13. The Relation of Resistance to Length and Area. — 

Experiment shows that the electrical resistance of a 
conductor varies directly with its length, the kind of 
material, and inversely with its cross sectional area. 
Analogy to the flow of water in iron pipes teaches us the 
same relation. The longer the pipe, the greater the resis- 
tance to flow ; the rougher or the more crooked the pipe, 
the greater the resistance ; the larger the pipe, also, the 
less the resistance, and the greater the flow. Stating these 
relations by means of a formula, we have 

R = K L (xs) 

in which R is the resistance, / the length, a the area of 
cross section, and K is a constant depending on the 
material of the wire, and means the resistance in ohms 
of unit dimensions of the material. If the length is 
expressed in feet and the area in circular mils, as it is 
usually in estimating the resistance of wires, then K 
is the resistance of a mil-foot ; that is, the resistance of 
a piece 1 foot long and 1 mil in diameter, or 1 circular 
mil in cross sectional area. A mil is the T oV °f an mcn > 
and circular mils are obtained by simply squaring the 
diameter in mils. Square mils are then obtained, if neces- 
sary, by multiplying circular mils by 0.7854. 



GENERAL LAWS OF RESISTANCE. 23 

In comparing two wires in order to obtain any one of 
the four quantities in (15), it will be more convenient to 
write down the formula for each wire, using subscripts to 
denote wire No. 1 and wire No. 2, then after substituting 
the given terms in each case, take the ratio of the two 
equations. For example, for wire No. 1 : 

R x = K x — (1), and for wire No. 2 likewise, 

R 2 = K 2 — (2). Or if areas are to be expressed in cir- 
cular mils as is usually better, 

R x = K x -^ (1) for wire No. 1, and 
a x 

R 2 = K 2 -^ (2) for wire No. 2. Now taking the ratio 

a 2 



of 


(0 


and 


co. 


we 


have 






















a 

^ 




X 


h 


X 


d? 



(3)- 

Example. — A length of 1000 feet of wire 95 mils in 
diameter has a resistance of 1 . 1 5 ohms ; what is the diam- 
eter of a wire of the same material whose resistance is 
5.045 ohms for 500 feet? 

Solution. — R x = 1.15 ohms; R 2 = 5.045 ohms. 
l x = 1000 ft. ; l 2 = 500 ft. 
K x = K 2 \ d x = 95 mils ; d 2 = ? 
Using (3) above, and making the proper substitutions, 



1. 15 1000 K 2 ^ 



X — X == ■ Whence, transposing, 
5.045 500 ^ 2 ^ 

d£ = — — = 1028.617 cir. mils, and 

2 X 5.045 

d 2 = V1028.617 = 32 mils. 



24 ELECTRICAL AND MAGNETIC CALCULATIONS. 

Example. — Find the resistance of 500 yards of cop- 
per wire 165 mils diameter, the resistance of 1 mile, 230 
mils diameter, being 1 ohm. — Day. 

Solution. — R 1 = ? ; R 2 = 1 ohm, 
4 = 500 yards; / 2 = 1760 yards; ^=165 mils; ^ = 230 mils. 
As before, 



^ = ^y^y^ = ^ = 5°° y ^1 y i^_ 
R 2 " 4 ^i <*?' 1 " 1760 A JCi * 76?' 

Whence R x = 0.55 ohm. 

For the comparison of the properties of substances, and 
especially of electrolytes, K, the specific resistance, is gener- 
ally expressed in microhms per cubic centimeter, instead of 
in ohms per mil-foot. The microhm is the one-millionth of 
one ohm. 

Example. — The specific resistance of copper per cubic 
centimeter is 1.6 16 microhms; find the resistance of 10 
meters of this wire 2 millimeters in diameter. 

Solution. — R x = 1.6 16 microhms ; R 2 = ? 

7T 

4= 1 cm.; / 2 =iooo cm.; a 1 =isq.cm.;a 2 =- X( T %) 2 sq. cm. 

X 2 = JC 1 ; 7T = 3.1416. 

Using the same formula as before, 

1.616 1 K x ' IT 

R 2 " 1000 K x 100 

Whence 

1.616X100X1000 . , , 

R 2 = - = 51,438 microhms = 0.05 ohm. 



GENERAL LAWS OF RESISTANCE. 



25 



i. Table of Specific Resistances, Specific Gravities, 
and Specific Heats. 



Substances. 



H 5 

K . 

'J 3 

M U 

. w 

sa 
-1 






i I X 

< a > 

p6 m ■»■ 

? w z 

BU £ H 



« 3 £ 



'/3 

<: . 

„ a z 

a I h 



Steel 

Silver, annealed . 
Silver, hard drawn 
Copper, annealed 
Copper, hard drawn 
Aluminum, annealed 

Zinc 

Platinum, annealed 
Iron, annealed 
Lead, pressed. . 
German silver 
Mercury . . . 
2 silver, i platinum 
Brass, rolled . . 



1.521 

1.652 

1.616 

1.652 

2.945 

5.689 

9.158 

9.825 

19.850 

21.170 

99.740 

24.660 

5.805 



9.05 
9.82 
9.61 

9.83 

17.52 

33.83 

5447 

58.44 

118.05 

125.89 

572.10 

146.65 

34-54 



0.282 

0.379 

0.376 

0.320 

0.316 

0.093 

0.253 

0.76 

0.278 

0.408 

0.307 

0.490 

0.305 



7.8 
10.5 
10.4 
8.9 
8.78 
2.6 

7-i 
21.5 

7.8 

"•3 

8.5 

J3-59 
8.4 



0.0568 

0.0933 

0.2122 

0.0935 
0.0323 

0.1 1 30 

0.0315 

0.0946 

0.0333 

0.09 



Degrees C. 
O-IOO 
O-IOO 



O-IOO 
O-IOO 
O-IOO 

2O-5O 



From Everett * 



Mi- 
crohms 

PER 
CU. CM. 



Remarks. 



Carre's Carbons at 20 C. 
Gaudin's Carbons . . . 
Retort Carbon .... 



Graphite Carbon 



3>9^7 
8,500 

67,000 

2,400 

to 

42,000 



Carre Carbons give |-ohm for 
a cylinder 1 meter long 
and 1 centimeter in di- 
ameter. 



Insulators. 



Ohms. 



Temp. 


20° 


C. 


24° 


C. 


28° 


c. 


46° 


c. 


46° 


c. 



Author. 



Mica . . . 
Gutta-percha 
Shellac . . 
Ebonite . . 
Paraffin . , 



Glass 
Air . 



8.4 x 10 

4.5 X 



10 



15 



9.0 X 10 
2.8 x io 16 
3.4 X 10 



16 



Ayrton & Perry 
Latimer Clark 
Ayrton & Perry 
Ayrton & Perry 
Ayrton & Perry 



Greater than any of the above. 
Practically infinite. 



J. D. Everett, C.G.S. System of Units, p. 178. 



26 ELECTRICAL AND MAGNETIC CALCULATIONS. 

14. The Relation of Resistance to Weight. — The weight 
of a body is obtained by multiplying its length by its 
cross sectional area, and this product by a constant repre- 
senting the weight of unit volume, and called the specific 
gravity of the material. For copper the weight per cubic 
inch is 0.32 lb., and per cubic centimeter 8.9 grams. 

For given lengths, the resistances of conductors vary in- 
versely with their weights. 

Example. — Determine the resistance of 100 lbs. of 
copper wire of a certain length, when another of the same 
length weighing 500 lbs. has 16.9 ohms resistance. 

Solution. — — - = — 4 • 

R 2 W x 

Substituting given values, 

Ri = 500 . 

16.9 100 

Tun r^ S°° X I ^'9 o 1. 

Whence R x = = 84. c ohms. 

1 100 ° 

Example. — Given that one mile of copper weighing 
19.74 lbs. has a resistance of 42.38 ohms, to find the re- 
sistance of a mile of copper wire having a diameter of 3 
millimeters. 

Solution. — 1 mile = 5280 X 12 = 63,360 in., 
3 mm. = 0.3 cm. = 0.3 x f = 0.12 ku, 

W 2 = 63,360 X 0.12 2 X 0.7854 X 0.32 = 229.3 lbs. 

As before, \ = W t ' 

Substituting known terms, 

42.38 _. 22 9-3 . 
R 2 19.74 ' 

Whence R 2 = 0.086 X 42.38 = 3.65 ohms. 



GENERAL LAWS OF RESISTANCE. 2J 

Example. — If iooo feet of copper wire 64 mils in di- 
ameter weigh 12.41 lbs., and have a resistance of 2.58 
ohms, what will 1 mile of wire 100 mils in diameter 
weigh, and what will its resistance be ? 

Solution. — Weight varies as length and cross-section. 

Hence — -. = j X -~ • 



"2 



_ , . . JV 2 5280 100 

From which — = x -==■ = 12.80. 

12.41 IOOO 61" 

Whence W 2 = 12.89 X 12.41 = 159.96 lbs. 

Also f = f x §! = 5l8o x ^ = 2 . l6 . 
y?! i x a 2 * 1000 Too 

Whence R 2 = 2.16 X 2.58 = 5.573 ohms. 

Example. — Suppose the second wire in the last exam- 
ple were iron instead of copper ; what would be its weight 
and resistance ? 

Solution. — Weight per cu. in. of iron = 0.278 lb., 
while that of copper is 0.32 lb. Hence, the total weight 

in the case of iron would be 159.96 x — ^— = n8qi lbs 

320 ° y 

Specific resistance of iron = 9.825 microhms per cu. cm., 

while that of copper is 1.6 16. Hence the total resistance 

of the iron wire would be 5.573 x 9 ' *\ = 33.88 ohms. 

15. The Relation of Resistance to Temperature. — The 
resistance of most substances increases with rise of tem- 
perature, the most important exception being carbon, whose 
resistance decreases with rise of temperature. The resis- 



28 ELECTRICAL AND MAGNETIC CALCULATIONS. 

tance of an incandescent lamp carbon when giving light 
is only about one-half that when it is cold. The alloy, 
nianganin, consisting of 12 per cent manganese, 84 per 
cent copper, and about 4 per cent nickel, changes very 
little in resistance with changes of temperature, and is 
furthermore peculiar in that its resistance increases very 
slightly up to about 45 ° C, after which it decreases again. 
We express the change in resistance per ohm per de- 
gree by calling it the temperature coefficient of resistance. 
Thus if J? be the resistance at zero C, or any low tem- 
perature, a the temperature coefficient of resistance, / the 
total rise in temperature, then the increase in resistance 
will be J? X at, a*nd the resistance at the higher tempera- 
ture, JR t , will be expressed by 

R t = R (i+at). (16) 

Example. — Find the resistance of a copper wire at 50 
C. when its measured resistance at 15 C. was 10 ohms. 

Solution. — i? = J /? 15 =ioohms; a for copper = 0.00406. 
(Kennelly & Fessenden) ; ^ = 50°- 15° = 35° c - Hence 

approx., J? 5() = 10 (1 + 0.00406 x 35) = n.421 ohms. 

The knowledge of temperature coefficients of resistance 
is applied practically in the determination of both low and 
high temperatures. Obviously it is only necessary to 
measure the resistance of a coil of wire, say platinum, at 
the required low or high temperature ; then knowing the 
same at any ordinary temperature, and assuming the co- 
efficient to remain constant, the simple application of (16) 
will obtain the unknown temperature, /. 



GENERAL LAWS OF RESISTANCE. 29 

Example. — Determine the melting temperature of a 
certain alloy when a coil of platinum wire whose resis- 
tance at o° C. is 10 ohms has a resistance in the melting 
metal of 50.8 ohms. 

Solution. — Applying (16), 

and supplying the known terms, 

50.8 = 10 (1 + 0.0034 /) = 10 + 0.034 /. 

„ , . . 50.8 — 10 _ 

From which / = = 1200 C, approx. 

0.034 

Example. — It is required to determine the tempera- 
ture of liquefaction of a certain gas when the platinum 
coil used above has a resistance in the liquid gas of 2.18 
ohms, assuming that the temperature coefficient does not 
alter. 

Solution. — Since in this case the temperature be- 
comes lower, the negative sign must be used in the for- 
mula; in other words, the resistance at the required 
temperature is less than at o°. 

Hence 7? t =J? (i — at), 

and 2.18 = 10 (1 — 0.0034 X /) = 10 — 0.034/. 

Therefore / = — = 230 below zero C. 

— 0.034 

Or, if the positive sign is retained, 

and 2.18 =10(1 + 0.0034/). 



30 ELECTRICAL AND MAGNETIC CALCULATIONS. 

From which / = — = — 2 xo°, 

+ 0.034 * 

which also means 230 below zero. Either method may 
be used. 

The temperature coefficient of resistance of alloys is in 
general lower than the coefficients of their component 
metals. The following are noted : 

German silver (60% copper, 25.4% zinc, 14.6% nickel) is, 0.00036 

Platinum silver (1 platinum, 2 silver) is 0.00030 

Platinoid (German silver and a very little tungsten) is, 0.00022 

Ordinary German silver is 0.00044 

Carbon has a negative coefficient, possibly about . — 0.0003 

Mercury in glass has a coefficient of 0.0008769 

Platinum has a coefficient of 0.0034 

Iron has a coefficient of 0.0045 

Copper has a coefficient of 0.0042 

16. Conductance and Conductivity. — It is sometimes con- 
venient, if not necessary, to make use of the conductance of 
a circuit, and the conductivity of a material. The conduc- 
tance of a circuit is the reciprocal of its resistance. The 
conductivity of a material is the ratio, expressed in per cent, 
of its conducting power to the conducting power of a 
standard, often pure copper, whose conductivity is called 
1, or 100 per cent. 

Example. — If the resistance of 1 foot of pure copper 
wire weighing 1 grain be 0.2106 ohm, and the resistance 
of a piece of ordinary copper wire 3 feet long weighing 
3.45 grains is found to be 0.5782 ohm, how does the 
conductivity of the latter sample compare with that of 
pure copper ? — Day. 



GENERAL LAWS OF RESISTANCE. 3 1 

W 2 l 2 a 2 

Solution. — -=~ = - 

W x l x a x 

From which — = 7 X -^ • 

tf 2 ^i ^2 

Substituting this value for — in the general formula given 

#2 

in (3) section 13, we have, 



Its K a L LW t 

assume K 2 = K x , then, 

From which 
R 2 = R x A X-^= 0.2106 X^j X — = 0.5494 ohm. 

This would be the resistance of the second wire if it were 
pure copper. But its actual resistance is 0.5782 ohm. 
Therefore its conductivity is 

^4Q4 

n = gc; per cent. 

5782 ^ F 

Example. — A circuit consists of a battery whose re- 
sistance is 2 ohms in series with two resistances of 1 o ohms 
and 15 ohms respectively. ♦ Find the conductance of the 
circuit. 

Solution. — Since all parts are in series the total re- 
sistance is the sum of all parts. 

Hence ^=2+10+15 = 27 ohms. 

Therefore conductance c = ^y = 0.037. 



32 ELECTRICAL AND MAGNETIC CALCULATIONS. 

17. Compound Circuits. — A compound circuit is one 
with two or more branches, or shunts. Thus in Fig. 1 the 

a 





Fig. 1. 

branches a, b, and c are called shunts relative to each 
other. In other words, they are connected in parallel with 
each other at the junction points C and £>. 

Example. — Obtain the formula for the combined re- 
sistance of a and b. Also for a, b, and c. 

Solution. — Let E be the E.M.F. between C and D, 
and R x , R 2 , and R g , the resistances of a, b, and c, respec- 
tively. The current flowing in the branches will then be 

^1= -£T> 7 2=-rr> and/>=— -• 
The total current 

/ = / 1 + /, + /, = *(-L+-I- + -L). 

But I = — where R is the combined resistance. 
R 

Therefore f =£ (±- + ±- + ±-^ 

But — is the total conductance of the parallel paths, and 
R 

— + — + — is the sum of the respective conductances, 
R x R 2 R B 

and 7? = ~W + ~p~ + lp~ ' 

/C K x K % K % 



GENERAL LAWS OF RESISTANCE. 33 

Hence, the total conductance of parallel circuits is the sum 
of the individual conductances. Reducing to a common de- 
nominator, equating the numerators, and solving for R, 
we get 

R^R 2 = ^^2 + 1M?l f° r the first and second. 
Then R (R l + R 2 ) = R,R 2 . 

From which R = f 1 ^ 2 ■ (17) 

Ri~t R 2 

Hence the resistance of two circuits in parallel, or shunts, 
is equal to the product of the two separate resistances divided 
by their su?n. 

Also for all three paths, 

R 1 R 2 R 3 = RR 2 R 3 + RR l R 2 + RR X R Z . 
Whence R (R X R 2 + R X R 3 + R 2 R 3 ) = R^R*. 

Or in words, the combined resista?ice is equal to the product 
of the separate resistances divided by the sum of their pro- 
ducts taken two at a time. 

In (18), if the resistances are equal to each other 

7p 3 D 

R = — =r^ = — , or one-third of one resistance. Making 

3 l 3 
this rule general, if R 1 represent each resistance and n the 

number of such in parallel, then the combined resistance 

will be 

* = -< 09) 

n N 7 



34 ELECTRICAL AND MAGNETIC CALCULATIONS. 

Example. — Find the resistance of 10 incandescent 
lamps in parallel, when the resistance of each, hot, is 200 
ohms. 

Solution. — Using (19), R = - 2 T °o°- = 20 ohms. 

Example. — Calculate the resistance of 3 wires in 
parallel, the resistances being, respectively, 5, 10, and 15 
ohms. 

Solution. — Applying (18), 

5 X 10 x 15 

K = ■ ■ = 2.72 ohms. 

5 x 10 + 5 X 15 +10 X 15 

18. Current Intensity in Compound Circuits. — Example. 
— A current of 42 amperes flows in a circuit composed of 
3 branches of 5, 10, and 20 ohms resistance, respectively. 
Find the current intensity in each branch. 

Solution. — From Ohm's law currents are manifestly 
proportional inversely to resistances, and directly to con- 
ductances. The conductances of the branches are, 
respectively, \, y 1 ^, 2V tota ^ 2 7 o> through which the total 
current of 42 amperes flows. Therefore, 

2V : 42 :: \ : x 1 ; whence, x x = 24 amperes. 
2~ 7 q : 42 :: ^V : x i % -> whence, x 2 = 12 amperes. 
2*0- : 42 : : ■%■$ : x z ; whence, x 3 = 6 amperes. 

19. Resistance and Drop of Potential. — According to 
Ohm's law, E = IR, it is clear that potential difference, or 
electromotive force, is directly proportional to the resis- 
tance, the current remaining the same ; or, we obtain drop, 
or difference of potential, between two points by multiplying 
the resistance between the points by the current flowing 
in the circuit. 



GENERAL LAWS OF RESISTANCE. 



35 



Example. — A cell of battery is connected in series 
with the following resistances : a 5 ohm coil of wire, a 10 
ohm coil, and 3 coils in parallel whose individual resistances 
are 5,10, and 20 ohms, respectively. Make a diagram of 
the circuit, and find the drop of potential over each of 
the resistances, and also over the cell whose constants are 
2 volts of E.M.F., and \ ohm internal resistance. 

Solution. — The total E.M.F. of 2 volts is used up, so 
to speak, in the total resistance. In other words, 2 volts 
is the total drop over the total resistance in circuit. The 
proportion of the drop over each resistance is therefore 
the same as the proportion of the resistances to the whole 
resistance. 




B 



kAAAAA 

MA/WW 





20 
Fig. 2. 

The total resistance is 

R = r + #! + R % + X* 

Substituting, 

r> , 5 x 10 x 20 

^ = 0.5 + 5 + 10 + 



5x10 + 5x20+10x20 



= 18.36 ohms. 



Therefore, 






Ohms. Volts. 


Ohms. 


Volts. 


18.36 : 2 :: 


5 


: #|_$ 


18.36 : 2 :: 


10 


• ^2? 


18.36 : 2 :: 


o-5 


: * ; 


18.36 : 2 :: 


2.86 


I X§\ 



36 ELECTRICAL AND MAGNETIC CALCULATIONS. 



whence, x x = 0.544 volts, 
whence, x 2 = 1.088 volts, 
whence, x = 0.056 volts, 
whence, x z = 0.312 volts. 

A voltmeter applied to a cell when no circuit is con- 
nected to it measures the total E.M.F. of the cell. But 
if a circuit is at the same time attached to the cell the 
voltmeter measures the total drop over the resistance of 
the circuit. The difference between these two readings 
of the voltmeter gives the drop over the internal resistance 
of the cell under the given conditions. The same thing 
is true of the dynamo. 

Example. — What is the resistance of the battery 
whose total E.M.F. is 91 volts, if, when it is connected in 
series with three resistances of 6, 3, and 9 ohms, a volt- 
meter connected across the 6 ohm coil reads 3 volts ? 

Solution. — Making the proportions as above, 

Ohms. Volts. Ohms. Volts. 

6 : 3 : : 9 : x x ; whence, x x = 4 J volts. 
6 : 3 :: 3 : x 2 \ whence, x 2 = 1 \ volts. 

Total external drop = 3 + 4-^ + 1 2 = 9 volts. Therefore 
the internal drop = 0.5 volt. Hence, 

6 : 3 :: .# : 0.5 ; whence, x = 1 ohm, the in- 
ternal resistance of the battery. 

20. Original Problems. — 1 . A Bunsen cell has the 
specific resistance of its inner and outer portions made 
alike, each being 9 ohms per cubic centimeter. The 



GENERAL LAWS OF RESISTANCE. 37 

central electrode is a \ inch cylinder of electric light 
carbon. The outer element is a plate of zinc 6 inches 
wide bent into a hollow cylinder. What will be the 
resistance of the cell when the jar is filled 4 inches deep 
with the solutions ? r = 0.165 ohm. 

2. How long is a 68 mil wire whose resistance is 
40 ohms, when 2 miles of the same kind of wire 100 
mils diameter have a resistance of 1 1 ohms ? 

k = 3-3 6 3 miles - 

3. There are two conductors, one of 35 ohms resistance, 
1728 feet long, and 12 square millimeters cross sectional 
area, specific resistance, 7 ; the other, 14 ohms resistance, 
432 feet long, and 8 square millimeters cross section. 
What is the specific resistance of the latter ? — Sloane. 

K 2 =7.4 ohms. 

4. What must be the length in feet and diameter in 
mils of a German silver wire, specific resistance 125 ohms, 
so that it may have a total resistance of 1250 ohms, when 
500 feet of the same material of twice the diameter have 
a resistance of 156.25 ohms? l x = 1000 feet. 

// x =io mils. 

5. Find the resistance of 15 miles of iron wire 0.3 inch 
in diameter, having given that the resistance of one mil- 
foot of iron wire is 59.1 ohms. — Day. 

R 2 = 52 ohms. 

6. Find the specific resistance in microhms per cubic 
centimeter of a wire, having given that its specific re- 
sistance per mil-foot is 9.8 ohms. 

T ^ 9.8(0.001 Xf) 2 Xi7rX io 6 r . _ 

K = - — ^— = 1.60^ microhms. 

12x4 



38 ELECTRICAL AND MAGNETIC CALCULATIONS. 

7. Find the resistance of \ mile of aluminum wire 
0.1 15 inch diameter, specific resistance per mil-foot being 
17.52 ohms. Also find the specific resistance in microhms 
per cubic centimeter, and compare with table 1. 

R = 3.49 ohms. 
K= 2.866 mich. Table 
gives 2.945. 

8. If the resistance of a wire 3 meters long and weigh- 
ing 3 grams is 5.88 ohms, what is the specific resistance 
in microhms per cubic centimeter, its specific gravity- 
being 20.337? — Day, 

Solution. — 

a x x 3°° X 20.337 = 3 grams. 
Therefore 

a x = sq. cm. 

300 X 20.337 

1 2 

Whence 

ley d? I 2 

H a = ^ lXy 2 X^- 2 =5.88 X — X 



k d i 3°° 3°° X 20.337 

= 9.638 microhms. 

Since J^ x = K^ and / 2 = 1. 

9. What is the resistance of a wire 3 feet long weighing 
5 grains, when it is known that one of the same material 
10 feet long weighs 16 grains and has a resistance of 
0.156? 

Solution. — 

D /4V w x , / 3 V 16 

R * = ^1(7) x -jr^ = 0.156 X ( — 1 X — = 0.045 ohms. 



GENERAL LA WS OF RESLSTANCE. 39 

10. How many pounds of a certain size of copper wire 
must I purchase to have a length of 500 feet and a resis- 
tance of 0.818 ohm, when it is already known that 20 
pounds require 1012 feet and have a resistance of 1.63 
ohms ? W 2 = 9.7 lbs. 

11. What is the temperature coefficient of resistance 
of a certain metal when it is found that a coil of wire 
made of it and having 10 ohms resistance at 15 C, in- 
creases to 16.8 ohms when heated to 215 C? 

a = 0.0034. 

12. A standard resistance of German silver wire is 
marked " 100 ohms, right at 15 C." What is the cor- 
rection necessary when the resistance is used at 2 8° C. ? 

7?28 = 100.572 ohms. 
Correction, 0.00572 per ohm. 

13. It is found by experiment that a spool of platinoid 
wire at 25 C. has a resistance of 50.275 ohms. How 
should it be marked for a standard correct at o° C. ? 

R° = 50 ohms. 

14. Find the fusion temperature of a certain metal 
when a 50 ohm coil of platinum wire " right at o°C." 
has a resistance of 220 ohms in the molten metal. 

/= 1000 C. 

15. Two samples of copper wire were brought to be 
tested for conductivity, so a length of 20 feet was cut 
from each sample. The first weighed 150 grains and 
had a resistance of 0.613 ohm, the second weighed 164 
grains and had a resistance of 0.547 ohm. Find the 



40 ELECTRICAL AND MAGNETIC CALCULATIONS. 

conductivity of each sample. Pure copper has a resis- 
tance of 0.2106 ohm and weighs 1 grain for 1 foot in 
length. — Day. C x = 9 1 .6 °f . 

C 2 = 93-9%- 

16. The resistance of 1 mile of copper wire whose 
diameter was 0.065 mcn was found to be 15.73 ohms. 
The resistance of a wire of pure copper 1 foot long and 
0.00 1 inch in diameter is 9.94 ohms. Find the conduc- 
tivity of the first wire. — Day. C t = 78.96%. 

17. Find the total conductance of a circuit in which 
are a battery whose resistance is 4 ohms, a wire resis- 
tance of 2>h ohms, another group of 2 wires in parallel 
whose separate resistances are 5 ohms and 15 ohms 
respectively. Also, how many amperes will flow, the 
battery E.M.F. being 2 volts? c= 0.114 

/ = 0.228 ampere. 

18. How much current will flow through each of the 
two parallel resistances in 17 ? 

I x = 0.057 amp. in 15-ohm wire. 
I 2 = 0.1 7 1 amp. in 5-ohm wire. 

19. Find the drop of potential over the parallel portion 
of 17. .£ = 0.855 volt. 

20. A dynamo machine whose resistance is 0.25 ohm 
is connected by lead wires whose resistance is 0.25 ohm to 
40 incandescent lamps in parallel, each having a resis- 
tance of 220 ohms hot. How many volts E.M.F. must 
the dynamo generate to give each lamp \ ampere of 
current? E = 120 volts. 



General laws of resistance. 41 

21. When 4 wires of 2.5, 5, 10, and 20 ohms resis- 
tance are joined in parallel, how much current will flow 
in each when a battery whose E.M.F. is 19 volts and 
whose internal resistance is 5 ohms is connected to 
them ? I = 3 amperes. 

I ± = 1.6 



7 2 = 0.8 



> amperes. 



7 3 = 0.4 

7 4 = 0.2 j 

22. How many incandescent lamps, each requiring 
1 ampere and having a resistance of 50 ohms, can be put 
in parallel, or multiple, on a machine giving 60 volts and 
having an internal resistance of 0.02 ohm, when the lead 
wires have a resistance of 0.1 ohm ? n = 83 in parallel. 

23. A dynamo giving 580 volts, having a resistance of 
1 ohm, is connected to a circuit containing in parallel 80 
groups of lamps, each group having 5 lamps in series 
requiring \ ampere of current. Find the voltage between 
the lines at the lamps, and also of each lamp. Also find 
the resistance of the wires the drop of potential over 
which is 40 volts, and find the percentage of the dynamo 
power used in the lamps, and the percentage lost in the 
machine itself. E = 500 volts for each group. 

e =100 volts for each lamp. 
R = 1 ohm in line. 
Lamps get 86.2% of total power. 
Internal loss 6.9% of total power. 
Line loss 6.9% of total power. 

24. A dynamo machine when not connected to any 
circuit, but running under excitation at full speed, gives a 



42 ELECTRICAL AND MAGNETIC CALCULATIONS. 

pressure at the brushes of 120 volts, as measured by 
a Weston voltmeter. When the full load of 60 lamps is 
put on the machine, the voltmeter reads 113.8 volts while 
the ammeter in the circuit reads 32 amperes. The volt- 
meter is now taken to the center of the system of lamps 
where it reads 109.5 volts. Find the drop on the line 
and in the machine. Also find the resistance of the line, 
of the lamps (average), and of the machine. Also 
determine the ratio of external to total energy of the 
machine. Internal drop = 6.2 volts. 

Line drop = 4.3 volts. 

Total drop = 10.5 volts. 

Lamp E = 109.5 volts. 

Resistance of machine = 0.194 ohm. 

Resistance of line = 0.134 ohm. 

Resistance of lamps = 205.2 (average). 

Ratio of ex. to total energy = 94.8%. 



25. Device M requires 5 amperes at no volts. A 220 
volt circuit is available, and a 20 ohm rheostat capable of 

20 ohms 

" A 



\j- 



-2-20- 




M 



•X ohms 



Fig. 3. 



carrying 5 amperes is at hand to put in series with the 
device. Now what resistance as a shunt to the device 
will enable it to operate under normal current ? 



GENERAL LA WS OF RESISTANCE. 43 

Since between A and B only no volts are permissible, 
there must be a drop in the 20 ohms of 

220 — no = no volts. 

Therefore the resistance in the circuit between A and B 
is given by the proportion, 

no volts : 2 o ohms : : no volts : x ; whence x = 2 o ohms. 

But this is the combined resistance of circuit M and 
shunt, and by (17) 



^1 + ^2 



Let JR X be the resistance of circuit M ' = JLJA =22 ohms. 
Therefore, substituting values, 

22 ^2 . 



20 == i~B"' 



From which 



7? 2 = 220 ohms = shunt resistance. 

The shunt will carry 2ffi — 5 = 0.5 ampere. 

The same result may be accomplished by putting addi- 
tional resistance in series. In this case, since the total 
current is 5 amperes, the total resistance must be 

220 -r- 5 =44 ohms. 

This will require 44 — 20 = 24 ohms additional resistance 
in series capable of carrying 5 amperes, and is much better. 

26. Suppose a 500 volt street car line is called upon to 
supply stationary motors in a factory requiring only 220 
volts and 10 amperes of current. Suppose further that a 
regulator rheostat of 18 ohms resistance can be put in 



44 ELECTRICAL AND MAGNETIC CALCULATIONS. 

series with the motors. What resistance now must be in 
parallel with the motors ? Also what additional series 
resistance would cause the motors to work normally ? 

Shunt resistance = 39! ohms. 

Series resistance =10 ohms. 

The latter would be used in practice, if either. 



ELECTRICAL ENERGY, 45 



IV. 

ELECTRICAL ENERGY. 

21. Electrical and Mechanical Energy. — Energy, or the 
capacity for doing work, is measured by the same units as 
work. In the C.G.S. system, the unit is the erg, or the 
work of 1 dyne of force acting through 1 centimeter of 
space. In ordinary mechanical work, the unit is the foot- 
pound, or the work of 1 pound of force acting through 1 
foot of space. In all cases work and energy are expressed 
by the product of force and distance. For estimating 
activity, ox power, that is, the rate of doing work, the horse- 
power equal to 33,000 foot-pounds per minute is used as the 
unit. For some purposes also the kilogram-meter is used 
as the unit of mechanical energy, being the work of 1 K. 
acting through 1 m. of space. As a smaller unit the gram- 
centimeter is chosen. 

As the unit of electrical power the watt is used, and 
for larger values the kilowatt is often convenient. The 
capacity of dynamos is always expressed in the latter unit. 

Expressing electrical power in watts by P, the formula is 

P=EL (20) 

Making the substitutions for E and I according to the 
law of Ohm, P = I *j?^ nAP=2 &. 

Jv 

So that the formula for P can be used which is moat 
convenient. 



46 ELECTRICAL AND MAGNETIC CALCULATIONS. 

It is sometimes necessary to make use of the whole 
amount of electrical work in a certain time. Representing 
the time in seconds by /, and the joules, or volt-coulombs, 
by W, the rule for the whole number is, 

W=EIt. (21) 

Example. — Reduce to ergs the energy employed per 
second by no volts of E.M.F. sending 10 amperes of 
current through a circuit. 

Solution. — 

1 volt = io 8 C.G.S. units ; 1 ampere=io~ 1 . 
Therefore 1 watt = io 8 x io -1 = io 7 ergs per second. 

From (20), 

P = £f= no X 10 = 1 100 watts. 

1 100 watts X io 7 = 11 X io 9 ergs per second. 

Example. — How many ergs of work will be done by 
no volts and 5 amperes in half an hour ? 

Solution. — * 

W=EIt= no X 5 X 1800 X io 7 = 99 X io 11 ergs. 

Example. — How many watts of electrical energy are 
equivalent to 1 H.P. of mechanical energy. 

Solution. — 

1 watt = io 7 ergs; 1 H.P. = 550 ft.-lbs. per second. 
1 m.= 3.28 feet; iK,= 2.2 lbs.; 1 K.-m. = 3.28 X 2.2 
= 7.23 ft.-lbs. 



Therefore 



_ 55° _ 



1 H.P. = ^— = 76.072 K.-m. per second. 
7.23 

= 76.072 x 1000 X 100 

= 76.072 x iq5 gr.-cm. per second. 



ELECTRICAL ENERGY. 47 

The acceleration of gravity is approximately 980 cms. 
per second, which is the ratio of the gravitation or practical 
system of units to the C.G.S. system. Hence the value 
given above is in C.G.S. units. 

76.072 X io 5 x 980 = 745.5056 X io 7 dyne-cm., or ergs. 
Therefore in round numbers 

746 watts = 1 H.P. (22) 

22. Electrical and Heat Energy. — No transformation of 
energy can take place without more or less waste in the 
form of heat, or unavailable energy. In every process of 
producing electrical energy by transformation from me- 
chanical or chemical energy, a considerable portion is lost 
as heat in the generating device. Its transmission to the 
points of utilization is also attended by much loss in heat- 
ing the conductors. Again, whether motors are used for 
power, or lamps for lighting, there is a still further trans- 
formation into heat ; in fact, being necessary in the latter 
in order to produce light. Hence it becomes very im- 
portant to understand the exact relations of electricity 
to heat energy. 

It has already been shown that electrical power is rep- 

resented by I 2 R, — and EI. The heat into which this 

power may be transformed is therefore directly porpor- 
tional to these quantities. 

Example. — Compare the heat produced in the con- 
ductors in transmitting 50 amperes through a resistance of 
1 o ohms with that when 2 o amperes are sent through the 
same circuit. 



48 ELECTRICAL AND MAGNETIC CALCULATIONS, 

Solution. — H x oo I* R v 
and H 2 oo 7 2 2 R 2 . But R x = R 2 . 

Hence 7^oo 50 2 x 10 = 25,000; 

and Hi 00 20 2 X 10 = 4000. 

Therefore § = 2 -*^ = 6*. 

-£/ 2 4OOO 

That is while the current in the first case is only 2 \ times 
that in the second, the heat produced is 65 times as great. 
Therefore, the resistance remai?iing unchanged, the heat varies 
as the square of the current. 

Example. — How will the line loss in carrying 10 
amperes through a 5 ohm circuit compare with that when 
the circuit has only 2 ohms resistance ? 

Solution. — H x 00 Ii 2 R v and H 2 00 I£R 2 - But 7[ = I v 
Hence H x io 2 X5 x 

H 2 "T^^" 22 ' 

That is, Heat varies directly as the resistance when the cur- 
rent is the same. 

Example. — The voltage of the dynamo is 116, of the 
lamps no, the line resistance is 10 ohms. How will the 
line loss compare with the loss if the resistance were only 
5 ohms, other conditions unchanged ? 

Solution. — Line drop =116— 110 = 6 volts. Then 

El (^ 
H x ~R X 10 1 
~H 2 - &?" 6 2 "" 2 ' 

£2 5 



ELECTRICAL ENERGY. 49 

Example. — When the line drop is 5 volts and the cur- 
rent 10 amperes, how much more heat is produced in 
transmission than in a circuit carrying 10 amperes, and 
having a drop of 2 volts ? 



Solution. — 

LE, 10 X c 

= 2f times. 



H x __ I X E X _ 10X5 _ 



H 2 I 2 E 2 10 x 2 

To measure absolutely the amount of heat certain units 
and constants are necessary. Joule* first found by experi- 
ment that 772! foot-pounds of mechanical energy will 
raise 1 pound of water i° Fahre?iheit, which he chose to call 
a unit of heat; 772 is then "Joule's Equivalent," the 
amount of mechanical work equivalent to 1 unit of heat. 
This would be 1390.59 ft.-lbs. per degree C, or 423.85 
kilogram-meters per kilogram-degree of heat. Later 
experiments made in the U. S. give the value of the equiv- 
alent as 427.52 K.-m. If the gram- degree of water is 
used as the unit, 427.52 gram-meters are equivalent to 
the small calorie. Representing the mechanical equiv- 
alent by J and reducing to ergs, 

/= 427.52 x 100 x 980 = 4.19 x io 7 ergs. 
1 calorie (water-gram-degree) = 4.19 x io 7 ergs of work. 

Example. — To find the ratio between joules of electri- 
cal energy and calories of heat energy. 

* British Association, 1843 ; also Phil. Mag., Vol. XXXII., 1843. See The Sec- 
ond Law of Thermodynamics in Harpers' " Scientific Memoirs," p. 111. 
t More recent researches fix this at 778 foot-pounds. 



50 ELECTRICAL AND MAGNETIC CALCULATIONS. 

Solution. — 

Ergs = watts x io 7 = IE x io 7 = io 7 ergs in i watt, 
i calorie = 4.19 x io 7 ergs. 



The ratio desired is then 

watt io 7 1 



= 0.24. 



calorie 4.19 x io 7 4.19 

Therefore, since the number of units is inversely as the 
size of the unit the number of calories corresponding to 
any number of watts, IE, will be 

H (calories) = ZSx 0.24. (23) 

To obtain calories multiply watts or joules by 0.24. 

The specific heat of a substance is the amount of heat 
expressed in calories necessary to change the tempera- 
ture of 1 gram of it i° C. Water is the standard and has 
a specific heat of one. 

Example. — Suppose no heat lost by radiation, how 
hot will a copper conductor 1000 feet long and 0.25 inch 
diameter and carrying iooo.amperes become in one sec- 
ond ? In 1 o minutes ? 

Solution. — R = K—„ = 9.8 ^ = 0.1568 ohm. 

d 2 y 250 2 

The volume= 1000 x 12 x 0.25 x 0.7854 = 589.04 cu. m. 

The weight = 589.04 X 145.45 (wt. per cu. in.^85,675.9 

grams. 

2 

Zf = 1000 X 0.1568 X 0.24 

= 37,632 calories per second. 
Heat acquired by the wire = mass X sp. heat X / (rise). 
jy = 85,675.9 X 0.0933 X /= 37> 6 3 2 - 



ELECTRICAL ENERGY. 5 I 

Therefore / = r 37? 32 = 4.7 above air. 

8 5> 6 75-9 X °-°933 
In 10 min. this would be 

4.7°X 600 = 2820 if possible. 
Copper melts at 1050 C. 

Although it is not known definitely because of so many 
influencing conditions, yet approximately X oVo of a heat 
unit is radiated from each square centimeter of external sur- 
face of a conductor for each degree rise of temperature above 
the air. 

Example. — How hot will the conductor in the last 
problem become, allowing the above rate of radiation to 
take place ? 

Solution. — 

Surface s of wire = (1000 X 12 x fXo.25 xf) X tt 
= 5 8 >9°5 sq. cm. 

Now the temperature will continue to rise until the rate 
of radiation is equal to the rate of development of heat, 
when it will remain stationary. 

Rate of production of heat = I 2 R X 0.24 = iooo 2 X 

0.157 X 0.24 = 37,680 calories per second. 
Rate of radiation of heat = x-oVo" x 5 8 >9°5 x *> 

in which / is the rise of temperature. 
Therefore, 

58,905 xt X ^roVo = 37>68o calories per second. 

r™ c 37> 68 ° X 4000 o r- 

Therefore /= — — — = 2CC2 C. 

58,905 M 

This is sufficient to melt the wire. 



52 ELECTRICAL AND MAGNETIC CALCULATIONS. 

Example. — Making use of the principles already given, 
find the diameter of a piece of copper fuse wire for the 
protection of a street-car motor, in order to carry 40 am- 
peres maximum current, the specific resistance of copper 
per cubic centimeter being 1.6 16 microhms, and taking 
its fusing point at 1000 C. 

Solution. — Represent the diameter of the wire in 
centimeters by d, and take 1 centimeter as the length for 
purposes of the calculation, since the length does not 
materially affect the point of fusion. The resistance of 
this will be 

1.616 X 1 1.616 X 4 , 

R = —. -. = — - R - 9 ohms. 

io 6 X Id 2 X -J 

From (23) 

rr „ n 9 1.616 X 4 1 

H = I 2 R X 0.24 = 40 2 X — e -~ X 0.24 cal. 

IO X 7TU 

This is the heat developed per second in the fuse-wire. 
For the purposes of protection, the rate of radiation must 
not become equal to this till the temperature of fusion is 
reached, or 1000 C. The heat radiated per second at 
1000 C. is, for 1 cm. length, 

H = X ltd X 1000 . 

4000 

Therefore 

1 , — 2 1. 616 X 4 

X 7rd X 1000 = 40 X. £— -^X 0.24. 



4000 *io 6 X 7rd 2 

From which 

,_ — 2 1. 616 X 4 4°°° 
d* = 40 X s ^ X 0.24 X 

IO 6 X 7T 2 IOOO 



ELECTRICAL ENERGY. 53 

Hence — 2 , , 

,, 40 x 1. 010 X 0.000391 

a* = ^- = 0.0010196. 

1000 

Therefore 

3, 

d = V0.0010196 = 0.1 cm. = 40 mils = No. 18. 

It will be clear that the final expression for d 3 just 
above may be written so as to make a general rule ; for 
40 = I 2 , 1.6 1 6 = K, and 1000 = /. Therefore the for- 
mula for general application is 



3 _ I 2 X K X 0.000391 



( 2 4) 



From the foregoing principles tables of " safe carrying 
capacity " are worked out. Some allow higher tempera- 
tures as " safe " than others. From experiment and cal- 
culation Dr. A. E. Kennelly has arranged a table of 
carrying capacities, now generally accepted, which are 
about 20 per cent lower than those of the National Code. 

It can also be seen from what precedes that the heat 

increases with I 2 and with -— • Hence 

d z 

I 2 

Now 30 C. above the surrounding air is at least a safe 
limit, though some would allow higher temperatures. In 
order that this temperature be not exceeded, the following 
empirical formula may be used for finding 7", the safe 
current for a wire of a given diameter d ; 

2500 

* A convenient practical formula for resistance due to any temperature t is 



*<== 



__ 9.6 (1 + .004 1) 1 



t- d* 



54 ELECTRICAL AND MAGNETIC CALCULATIONS. 

23. Efficiency of Transformation and Transmission of 
Energy. — The efficiency of conversion of an electrical 
generator is the ratio of the total electrical energy gen- 
erated to the mechanical energy supplied to the machine. 
Representing the efficiency of conversion by C, 

E.H.P. (total) 

C = M.H.P. ' (26) 

The commercial efficiency of a generator is the ratio of 

the electrical energy supplied to the external circuit to 

the mechanical energy used in driving the machine. 

Therefore 

E.H.P. (external) 

C ° m - Eff ' = ALHJ\ ' (27) 

The electrical efficiency of a generator is the ratio of the 
external electrical energy to the total electrical energy 
developed in the armature of the machine. Hence, 

Elec. Eff. = E ^ p (e f 6r " al) • (28) 

E.H.P. (total) v ' 

Station efficiency is often employed to designate the 
ratio of the power delivered to the line terminals on the 
switchboard to the power supplied to the engine by the 
steam. The difference between these two quantities is 
the total loss in the station beginning at the engine. 

Plant efficiency may also be used to represent the ratio 
of the energy delivered to the lamp or motor terminals to 
the energy supplied to the steam engine. The latter is 
important in judging of the performance of the whole 
system and in fixing charges to consumers. 

The various losses in a plant are friction and radiation 
losses in the engine ; losses in belting and so forth ; fric- 



ELECTRICAL ENERGY. 55 

tion, hysteresis, eddy current, and I 2 R losses in dynamos, 
and I 2 R losses in the switchboard connections and in the 
line. Evidently that plant will be most efficient which 
has observed the means of reducing each of these losses 
to the lowest possible. 

Example. — What is the commercial efficiency of a 
dynamo, when a dynamometer shows that it absorbs 
8 H.P. of mechanical energy while furnishing 92 incan- 
descent lamps with 46 amperes of current at 1 1 5 volts ? 

Solution.— E.H.P. = II5 * 46 = 7. 

746 

Com. Eff. = -|- = 87.5%. 

Example. — The resistance of an armature is 0.02 ohm 
and the shunt fields have a resistance of 25 ohms. The 
generator absorbs 10 H.P. and delivers 50 amperes to the 
line, while the brush E.M.F. is 118.8 volts. What is 
the efficiency of conversion, commercial efficiency, and the 
electrical efficiency of the machine? Also the per cent 
of loss in field magnets and armature ? 

Solution. — External watts = 118.8 x 50 = 5940. 

1 18.8 
rield current = = 4.75 amperes; 

o 

Field watts = 118.8 x 4.75 = 564.3. 



Armature watts loss = 54-75 X 0.02 = 60, 

Therefore 

Total watts = 5940 + 564.3 + 60 = 6564.30 
10 H.P. = 7460 watts supplied. 



56 ELECTRICAL AND MAGNETIC CALCULATIONS. 

Hence 

Eff. Con., C = 6564.3 -T- 7460 = 87.99%. 

Com. Eff. = 5940 -r- 7460 = 79.62%. 

Elec. Eff. = 5940 -T- 6564.3 = 90.48%. 

Total machine loss = 60 + 564.3 = 624.3 watts. 

Electric loss = 624.3 "*" 6564.3 = 9.51%. 

These figures would show a very badly designed machine 
for modern times. 

Example. — Calculate the efficiency of a long distance 
line when 50 amperes at 3600 volts are supplied to it, its 
resistance being 10.8 ohms. 

Solution. — Watts supplied = 3600 x 50= 180,000. 
Line loss = 50 X 10.8 = 27,000 watts. 
Therefore, 

Watts delivered = 180,000 — 27,000 = 153,000. 
Hence Line eff. =^|^ = 8 5 %, 
and Line loss = ffififr = 15%. 

Example. — A plant consists of a generator of 94% 
efficiency, step-up transformer (lower to higher voltage) 
96% efficiency, line 90% efficiency, lowering or step-down 
transformer 95%. The engine that drives the dynamo 
has an efficiency of %%°/ - If 339-3 2 K- w - of energy is 
delivered at secondaries of step-down transformers, cal- 
culate the plant efficiency and the H.P. supplied to the 
engine ; also the station efficiency. 

Solution. — Let 100% = energy supplied to engine, 
0.88 x 0.94 X 0.96 X 0.90 X 0.95 = 67.89% = plant effi- 
ciency. 



ELECTRICAL ENERGY. 57 

Therefore 

67-89% = 339-3 2 K - w - 

i%=5- 

100% =500 K.W. supplied to engine. 

H.P. of engine = ^Vf — = 6 7°- 

The station efficiency = 88% of 94% = 82.72%, if we 
include the belt losses and slipping losses in the 12% en- 
gine loss. 

The efficiency of most machinery is much higher at full 
load than at very small loads. This is largely true of en- 
gines, transformers, dynamos, and motors. However, in 
the latter the difference between full-load efficiency and 
say one-half load is small. The efficiencies as previ- 
ously given are for full load. In engines the losses are 
almost constant for all loads. Hence an engine which at 
full load would have an efficiency, say of 90%, at half 
load would have only about 82^ efficiency. To illus- 
trate, take 100 H.P. supplied, 90 H.P. delivered, making 
10 H.P. lost. Half load delivers 45 H.P., requiring 45 + 
10 = 55 to be supplied. Efficiency = f 4 = 82 °/ . There- 
fore all engines should keep loaded as fully as possible. 
For dynamos, motors, and large transformers, the effi- 
ciency changes but little with change of load, if the 
load is not allowed to fall below about one-half. 

Example. — A street car with its load weighs 10 tons, 
and the average grade of the street is 5%. Allowing an 
efficiency for the motors and gearing of 75%, and a hori- 
zontal speed of 10 miles an hour, what power must be 
applied to the motor ? 



58 ELECTRICAL AND MAGNETIC CALCULATIONS. 

Solution. — The horizontal effort may be taken as 25 
lbs. per ton, making 10X25 = 250 lbs. for this car. 
Work done horizontally per minute is then 

TT _ 5280 X 10 X 250 _ „ 

W 1 = - — = 220,000 ft.-lbs. 

60 

The vertical lift per minute is 

5% of 5280 X 10 , 

^ ^ = 44 ft. per mm. 

DO 

Work done vertically per minute is therefore 

JV 2 = 10 X 2000 x 44 = 880,000 ft.-lbs. 

Total work required at the wheels of car 

= 220,000 + 880,000 = 1,100,000 ft.-lbs. 

Horse-power necessary to be taken from the line is 

1,100,000 TT _ 

-*- 33,000 = 44.4 H.P. 



0-7S 

Example. — What efficiency will be necessary in a sta- 
tionary motor in order that it may develop 450 volts coun- 
ter E.M.F. when the applied E.M.F. is 500 volts? 

Solution. — Since the output is the product of the cur- 
rent and the counter E.M.F., or IE 2J and the intake is the 
product of the current and the applied E.M.F., or IE V the 
efficiency is, 

TE 2 E 2 450 ^ 

Example. — Determine at what efficiency a motor 
gives its greatest output per minute. 



ELECTRICAL ENERGY. 59 

Solution. — Since the current through a motor is 
in which E 1 and E 2 are the applied and counter 



E t — E 2 



R 
E.M.F's. respectively, and R is its resistance, the output is 

77 (77 - 77 \ 

E 2 I = — — — x — — • In this E 1 and R are constants. 

R 

Obviously, then, the numerator, consisting of this product 
of a variable and the difference between a constant and 
the variable, will be the greatest when E 2 = \E X . In other 
words, the output is at the greatest rate when the counter 
E.M.F. is \ the applied E.M.F., and the efficiency, there- 
fore, 50%. To illustrate, 

Suppose the applied E.M.F., E 1 = 100 volts. 
Suppose the counter E.M.F., E 2 = 50 volts. 

Then the output will be represented by 

E 2 (E x - E 2 ) = 50 (100 - 50) = 2500. 

Let the speed increase, so that the counter E 2 = 60 volts. 
The output is now 

60 (100 — 60) = 2400. 

Again, let a decrease of speed reduce the counter 
E.M.F., E 2 , to 40 volts. Then the output is 

40 (100 — 40) = 2400. 

This shows that the greatest rate of work is done at an 
efficiency of 50 per cent, but is modified in practice. 

24. Original Problems. — 1. A voltmeter connected 
across a Buckeye lamp read 109.5 Y0 ^ ts when 0.6 ampere 
was flowing through the lamp. How many joules of 
energy are used by the lamp per second, and how many 



60 ELECTRICAL AND MAGNETIC CALCULATIONS. 

ergs ? How many such lamps can be maintained per 
H.P. ? How many calories of heat are developed per 
second in each lamp? If the lamp has a candle-power of 
14, what is the efficiency in watts per c.p. ? 

JV= 65.7 joules= 657,000,000 ergs= 15.7 cal. 

Eff. = 4.7 watts per c.p. 

1 1 lamps per H.P. 

2. I wish to supply 50 lamps at no volts, resistance of 
each being 220 ohms hot. The line loss is 5 °j , the com- 
mercial efficiency of the generator can be counted as 92 %, 
and the engine and belt losses are 15%; what I. H.P. 
will be necessary to operate the plant ? 

I.H.P. = 5 (strictly 4.9). 

3. Find the H.P. required to furnish 40 arc lamps in 
series with 7 amperes of current, the resistance of each 
lamp being 8 ohms, line 25 ohms. How many watts are 
necessary per lamp? Assuming a normal candle-power 
of 1200 for each lamp, find the efficiency in watts per c.p. 
Compare last result with the efficiency in 1. 

H.P. = 22.6. 

Eff. = 0.326 watt per c.p. 

4. An engine on full load indicates 12.5 H.P. Assum- 
ing engine and belt losses to be 12%, and the commer- 
cial efficiency of the dynamo 90%, line loss 5%, how 
many no volt incandescent lamps constitute the load? 
Also, what is the voltage at the brushes ? How many in- 
candescent arcs could be supplied at 4 amperes each ? 

No. lamps, n = 127. 
Brush volts = 115.8. 
Inc, arcs, n = 16. 



ELECTRICAL ENERGY. 6 1 

5. The resistance of a certain dynamo machine's arma- 
ture is 0.016 ohm, that of the external circuit 0.757 ohm. 
The power required to operate the machine is 7.604 H.P., 
and the current produced is 83.7 amperes. What are the 
efficiency of conversion and the commercial efficiency of 
the machine ? C = 95.5 °/ . 

Com. Eff. = 93.5%. 

6. A circuit consists of 100 incandescent lamps 
arranged in 20 parallel groups, each group containing 5 
lamps in series. The voltage between the main wires at 
the center of the lamp system is 550, and the resistance 
of each lamp is 212 ohms, average. Find the watts and 
ergs used per lamp. Also the amount of current flowing, 
and the efficiency of transmission if the line resistance is 
8 ohms. 

Watts per lamp = 57 = 57 x io 7 ergs per sec. 
Eff. of trans. = 87%. 
/= 10.37 amperes. 

7. If the following set of conditions are representative, 
how many no volt incandescent lamps can be supplied 
per I.H.P. of the engine ? Lamps 100, drop in line 
6.5 volts, resistance of fields, shunt wound, 25 ohms, of 
armature 0.025 onm 3 ^ or generator 96%, engine and 
belt losses 15% ? Find the plant efficiency. 

Lamps per I.H.P. = 9. 
Plant eff. = 73%- 

8. Compare the external heat with the internal in the 
following cases ; also determine the efficiency in each 
case ; the battery resistance is equal to 7 feet of a certain 
size of wire, and for the first case has its terminals con- 



2 ^ 

= 22.2% 



62 ELECTRICAL AND MAGNETIC CALCULATIONS. 

nected through 2 feet of the wire ; for the second case 
the battery terminals are connected through 53 feet of 
the wire. , N Zf (ex.) 2 „ 

^w^rr eff - = 2 +7 

V ' -^(m.) 7 53+7 6/ 

9. A current of 0.75 ampere is sent for 5 minutes 
through a column of mercury whose resistance is 0.47 
ohm, and which weighs 20.25 grams, and has a specific 
heat of 0.0332. Find the rise of temperature, assuming 
no heat lost by radiation. — Day. t = 28 C. 

10. Assuming the rate of radiation previously given, 
find how hot an iron wire \ inch in diameter and 100 
feet long will become in the air when it carries 20 am- 
peres. / = 47 . 

11. What must be the diameter of a German silver 
wire 50 feet long to carry 5 amperes so that it may not 
rise to exceed 30 above the temperature of the sur- 
rounding air? d = 74 mils = No. 13. 

12. How many amperes of current may a copper wire 
64 mils diameter and 500 feet long carry so as not to 
exceed 25 above the temperature of the air ? 

/= 13 amperes. 

13. A coil of platinum wire is immersed in a tank 
holding 3 quarts of water which is constantly stirred 
while 4 amperes of current pass through the wire which 
has a length of 20 feet, and diameter of 40 mils. The 
initial temperature of the water is 15 . Assuming no 
loss by radiation, what will be the temperature of the 
water after 10 minutes? /= 15.56° C. 



ELECTRICAL ENERGY. 63 

14. Find the gauge of a zinc wire to be inserted in a 
circuit to carry 250 amperes as a maximum current, when 
zinc fuses at 422 C. d = 270 mils = No. 2. 

15. What size of copper wire must be used for the 
protecting fuse of a circuit designed to carry normally 
300 amperes, allowing temporary over-loading of 40% ? 

d = 0.48 cm. = 188 mils = No. 5. 

16. What is the fusion temperature of an alloy when 
a piece of gauge number 16 fuses at 20 amperes, know- 
ing that the specific resistance is 24.66 ? 

/= 1800 C. 

17. If a certain machine when run at 2200 r.p.m. 
generates 116 volts at the brushes, and supplies no volts 
and 22 amperes of current at a certain distance, how 
many times as far will it furnish the same power when 
run at 2300 r.p.m. ? 1.9 times as far. 

18. The poles of a dynamo whose resistance is 0.08 
ohm are in the first place joined through 4 incandescent 
lamps in series, in the second place through 4 in parallel. 
The resistance of each lamp is 50 ohms, and the 
mechanical energy supplied to the machine is the same 
in both cases. Compare the amounts of heat developed 
in the machine ? — Day. The lamp resistance is here 
wrongly assumed constant. H 2 16 

j2i~T" 

19. A circuit contains 20 incandescent 220 ohm lamps 
in parallel, the lead wires having a resistance of 1 ohm 
and the machine 0.1 ohm. Now a wire whose resistance 
is 0.5 ohm is connected across the leads at the lamps, 



64 ELECTRICAL AND MAGNETIC CALCULATIONS. 

while the E.M.F. at the brushes is 120. Compare the 
current in the lamps with that in the wire. How does 
the heat in the machine compare in each case with that 
in the external circuit ? How much greater load is 
placed on the engine by the wire ? 

Normal /= 10 amperes. 
With wire, /= 81 amperes, 3.5 in lamps. 
Normal heat in machine H x = 10 watts. 
2d heat in machine H 2 = 656 watts. 
Load with wire 8 times normal. 

20. A dynamo and a motor are exactly alike, each 

having a normal E.M.F. of no volts when running at 

2000 r.p.m. The dynamo is kept running at the normal 

speed, but the load of the motor is such that it runs at 

1500 revolutions per minute. What must be its efficiency 

at this load and speed ? 

Counter E.M.F = 82.5 volts. 

Eff. = 75%. 

21. What H.P. must a hoisting motor be capable of 
developing when 900 pounds are to be lifted 80 feet per 
minute, the cage being counterbalanced by a weight 
suspended from the drum, and making an allowance of 
50% of its power for losses due to friction in all parts ? 

H.P. = 4.37. 

22. What H.P. would be required if the cage were 
drawn at the same speed up an incline in which there is 
a rise of 10 feet for every 10 feet in a horizontal direction, 
in other words when the grade is 100% ? 



ELECTRICAL ENERGY. 65 

Solution. — Here the ratio of the vertical lift to the 
slope must be taken as the ratio of power required to 
that for a vertical lift. Slope = Vio 2 + io 2 = 14.14. 

Therefore the power will be 4.37 X = x H.P. 

14.14 

23. A motor is to operate a pump lifting 20,000 gallons 
per hour 500 feet high. Allowing an efficiency of 60% 
in the pumping system, and given that 1 gallon weighs 
8.33 lbs., how many H.P. must the motor develop ? 
Also putting its efficiency at 90% and operating it on a 
220 volt circuit, how many amperes of current will it 
require? 70 = H.P. developed. 

78 = H.P. supplied. 
/= 265 amperes. 

24. An electric car with its load weighs 12 tons and is 
to maintain a speed of 10 miles an hour on a grade of 
2%; that is, 2 feet rise for every 100 measured hori- 
zontally. Allowing a traction effort of 25 lbs. per ton, 
how much power is necessary at the axles, and assuming 
an efficiency of 75% in motor and gearing, how many 
amperes will it take from a 500 volt circuit ? 

Power at axles = 21 H.P. 
/= 41.8 amperes. 



66 ELECTRICAL AND MAGNETIC CALCULATIONS. 



WIRING FOR LIGHT AND POWER. 

25. Drop of Potential and Size of Leads. — The simplest 
case of wiring is that for arc lamps which are all in series, 
70 to 80 volts being required for each lamp. The ma- 
chine at the station automatically changes the pressure to 
suit the number of lamps, while the current is kept con- 
stant at 5 to 7 amperes, depending on the system used. 
This is therefore a constant current system. 

The resistance of the wire is generally taken so that 
not to exceed 10% of the dynamo volts will be required 
to pass the current through it. 

Example. — A certain circuit carries 40 T.H. arc lamps 

in series, inclosed type ; each lamp requires 75 volts, the 

current being 5.5 amperes. Allowing a line drop of 5% 

what must be the difference of potential at the brushes, 

and what size of wire must be used, its length being 

3 miles ? 

E = 3157.8 volts. 

R = 28.69 ohms. 

Wire is No. 12, A.W.G. or B. & S. 

Incandescent arc lamps are those designed to work in 
parallel on the same circuit as incandescent lamps. They 
are used for both interior and exterior lighting, and wound 
for no or 50 volts. The calculations are the same in 



WIRING FOR LIGHT AND POWER. 6 J 

this case as for incandescent work, and so will not be 
treated separately. 

Although for incandescent lighting and power on the 
parallel, or constant potential system, the wiring often 
becomes complex and the exact distribution of copper in 
the various parts of the net-work a matter for much study, 
yet the principles which govern the calculations are the 
same as for simple circuits, and are all based on Ohm's 
law. The problem may be to calculate the leads and 
lamp circuits leading from street mains or transformers to 
the lamps, or it may include the mains and all the wiring 
of a complex system. 

Example. — The street mains are kept at the constant 
potential of 116 volts. Lamp leads are run out to the 
center of a system of 150 incandescent lamps. The volt- 
age at the point from which the lamp circuits leave the 
leads is 112. Calculate the resistance per foot of the 
wire to be used if the leads are 100 feet long, and obtain 
from the table the corresponding gauge number. 

Solution. — Line drop = 116—112 = 4 volts. 

/for 150 lamps = 150 X \ = 75 amperes. 

E 4 
Hence R, from Ohm's law, = — = — = 0.0a ohm. 

^ 75 M 

Leads contain 200 feet of wire ; hence the resistance per 
foot of the wire to be used is 

R = 0.053 -T- 200 = 0.000266 ohm ; and per 1000 feet, 
R = 0.000266 X 1000 = 0.266. From the table this 
corresponds to No. 4 A.W.G. 



68 ELECTRICAL AND MAGNETIC CALCULATIONS. 



Example. — Determine the size of the wire for the 
lamp circuits in the above problem assuming all to be of 
equal length, averaging 30 feet from the junction with 
leads, and carrying 10 lamps each. 

Solution. — Drop in lamp circuits =112 — 110 = 2 volts. 

Current in each = 10 X -J- = 5 amperes. 

0.4 ohm. 



R of each = — = - 

1 5 



Length of wire in each circuit = 2 x 30 = 60 ft. 
Hence per 1000 feet, 

R = 0.4 X 10 00. _ 6.66 ohms = No. 18 A.W.G. 

The rules for correct wiring do not permit wire smaller 
than No. 14 to be used. If 14 is used here the resist- 
ance for 60 feet = 0.158 ohm, and the drop will become 
E = 0.158 X 5 =0.79 volt. This means that the lamps 
will burn at 112 — 0.79 = 111.2 volts, or else the volt- 
age at the street mains must be reduced about 1 volt so 
that the lamps will burn at no volts. 

50 I'ps. 



\f 



<— H-5-> 



200 ft. 



200 ft. 



300 ft. 



1.6 vt. drop 



1.6 vt. drop 



a 50 I'ps. 
_ 



50 I'ps 



Fig. 4- 



Example. — Let the diagram, Fig. 4. represent the 
conditions of a circuit, to solve for the size of the various 
portions of the leads. 



WIRING FOR LIGHT AND POWER. 69 

Solution. — The current required by the first group 

of 50 lamps, each of which has a resistance of 220 ohms, 

will be 

t I][ 5 — x -6 
J x = — — = 2C.7 amperes. 

1 220 -r- 50 D ' * 

__ , T 11^.4 — 1.6 

The second, A = — = 2 s. 41 amperes. 

2 220-5-50 D ^ v 

m . . . , ^ 1 1 1.8 — 1.8 

The third, A = = 2 c amperes. 

220-^50 ° r 

Therefore the total current flowing through the first 
section of the leads is 

I' =25.7 + 25.41 + 25 = 76.1 amperes. 

Total current flowing through the second, 

I" = 25.41 + 25 = 50.41 amperes. 

In the last section, 

I ,n = 25 amperes, 

Hence the resistance of the first section is 

j? E I - 6 i. 

R, = — = — — = 0.021 ohm. 

I 76.1 

This is for 200 X 2 = 400 feet of wire. 
The resistance for 1000 feet then is 

_ 0.021 X 1000 , 

R = = 0.0C2 ohm. 

400 ° 

Table No. 2 gives this to be about No. 000 wire. 

The resistance of the second section is 

Ro = — - — = 0.0317 ohm for 400 feet. 
2 50-41 6 ' 



7o 



ELECTRICAL AND MAGNETIC CALCULATIONS. 



2. Table of Wiring Constants. 



6 


■ 




Wt., 


Lbs. per 




Safe Carrying 
Capacity.— 


w 
o 
p 


►J 


Area 


< IOOO FT. 


7?, Ohms 


A 


MPERES. 




Insulated. 


H 
04 


w 


w 




W 
H 
W 


Arc 
Mils. 






per 

IOOO FT. 


& 


64 


P4 


1 br. 

04 £ 


s> 


n 






Bare. 


§8 

X 04 

w 



a . 

H 

fc 






w 

in 

p 




H 



<5 










* 


£U 




pq 


H 


§ 


oooo 


460 


2II,6oo 


640.5 


730 




•05075 


620 


175 




000 


410 


167,805 


508.O 


588 




.064OO 


525 


147 




oo 


365 


I33»079 


402.8 


476 




.08070 


538 


124 




o 


325 


!05>593 


319.6 


385 




.IOI7I 


369 


104 




i 


289 


83>695 


2534 


308 




.12832 


309 


87 




2 


258 


66,373 


201.0 


250 




.16181 


260 


73 




3 


229 


52*634 


J 59-3 


200 




.20404 


219 


62 




4 


204 


4i*743 


126.4 


l6o 




.25729 


183 


49 




5 


182 


33* 102 


100.2 


*33 




.32444 


154 


44 




6 


162 


26,251 


79.46 


in 




.40914 


130 


37 




7 


144 


20,817 


63.01 






•5 X 593 


109 


3i 




8 


128 


16,510 


49.98 


72 




.65053 


92 


26 




9 


114 


13,094 


39-64 






.82022 


77 


22 




10 


102 


10,382 


31.43 


5 1 




1.03458 


65 


18 




ii 


91 


8,234 


24.93 






1.30434 


54 


154 




12 


81 


6*53° 


19-77 


33 


23.8 


1.64474 


46 


x 3 


6-5 


13 


72 


5*i78 


15.68 




18.2 


2.07400 


38 


11 


5-2 


14 


64 


4,107 


12.43 


25 


14.7 


2.68044 


32 


9^ 


4.1 


15 


57 


3*257 


9.86 




11.5 


3.29782 


27 


7.6 


H 


16 


5i 


2*583 


7.82 


iS 


9-i 


4-15812 


23 


6.4 


2.6 


17 


45 


2,048 


6.20 




7.2 


5-24363 


J 9 


54 


2.0 


18 


40 


1,624 


4.92 


11 


5-7 


6.61208 


16 


4-5 


1.6 


19 


35 


1,252 


3-79 




4-5 


8.562 






13 


20 


32 


1,022 


3-09 




3^ 


10.498 






1.0 


21 


28 


810 


2.45 




2.8 


13.237 






.8 


22 


25 


642 


1.94 




2.2 


16.690 






.6 


23 


23 


509 


1.54 




1.8 


21.050 






•5 


24 


20 


404 


1.22 




1.4 


26.547 






4 



WIRING FOR LIGHT AND POWER. 7 1 

Per 1000 ft., 

n 0.0317 X 1000 . XT 

R = — — - = 0.079 ohm, or No. 00 wire. 

400 

The resistance of the third section is 

1 8 

R s = — = 0.072 ohm for 600 feet. 

2 5 
Per 1000 ft., 

_ 0.072 X 1000 , AT 

R = — I = 0.120 ohm, or No. 1 wire. 

600 

The practice in wiring is so to distribute the drop of 
potential, by adjusting the resistance of the circuits, that 
it decreases from the dynamo outwards to the lamps. 
For example, let a small plant supply a circuit so as to 
lose 6 volts. We may wire for 3 volts drop in the mains, 
2 volts in the leads, and 1 volt in the lamp circuits. 

26. Wiring Formulae and Diagrams. — For greater con- 
venience in calculation we may apply a certain formula 
for finding the number of circular mils required in a 
wire to supply a given number of lamps, at any distance 
required, for a given drop of potential. 

Example. — Let it be required to derive the formula 
for the circular mils required in any portion of a circuit 
when the distance and the number of lamps and per cent 
of drop are given. 

Solution. — We have already for resistance (.15), 

R = kL = io . 79 J_. 

However, 10.79 * s nere use d as more nearly represent- 
ing the specific resistance of ordinary commercial copper 



72 ELECTRICAL AND MAGNETIC CALCULATIONS. 

than 9.8, the specific resistance of pure copper which was 
used previously in this book. By transposition of the 
above formula we have for circular mils, 

We also have for the resistance of any number of lamps 
n in parallel when the resistance of each one is r, 

r (hot) 

'!=— " Z . (30) 

Since resistances are proportional to the drops of 
E.M.F. in them, 

R (line) °f drop in line . 

r 1 (lamps) "" 100 — % * ^ 3 ^ 

Here 100— °f drop in line = per cent drop in lamps. 
The per cent is calculated on the highest voltage of the 
system as a base. By transposition of (31), 

* = ^4- (33) 

IOO - % w ' 

Putting in (32) the value of r x from (30), 

^ = ^(hot) x _% (33) 

n 100 — °j 

Equation (33) gives the line resistance in terms of the 
resistance of one lamp hot, the number of lamps, and 
the per cent drop. Having thus the resistance and 
knowing the length of the wire, the size could be deter- 
mined as we have already done. But to make the work 
still more complete, place this value of R in (29) ; then 
for circular mils we have 



WIRING FOR LIGHT AND POWER. J$ 



d* = IO - 79/ 



'•(hot) % 



n ioo — °f 

o- rr • ^2 IO.79/ X^2 IOO — % / 

Simplifying, *■ _«j__ x — ^ (34) 

But / the length is twice the measured distance D ; that 
is, / = 2D. Hence, placing this in (34), 

# 21.58 x {Dy.fi) 100- % 

The product Dn is called the " lamp feet " of the circuit, 
being the product of the distance in feet by the number 
of lamps, r is 220 for no volt lamps and 50 for 50 volts. 

Example. — A circuit is to be run 400 feet for 150 
lamps at a loss of 5 per cent. The lamps are rated at 
no volts and \ ampere. Determine the size of the main 
wire. — Badt. 

Solution. — Applying directly (35), 

.. 21.58 X 400 X 150 100 — 5 • .1 

d 2 = — — X - = 111,823 cir. mils. 

220 5 J 

This corresponds to No. o wire A.W.G. or B. & S. 

It is evident that as long as 1 1 o volt lamps are used 

21. c8 tit 1 21. c8 .„ . 

— ^— , or if co volt lamps are used — ^— will be constant, 
220 ° r 50 

and the value of the fraction may be so used in the wiring 

formula. Then (35) will become for no volt lamps 

d 2 = 0.098 X Dn X IO ° ~ ^ - (36) 

% 

* Instead of 21.58 m this equation for carbon lamps, use the constant 12 for 
30-watt tungstens, constant 16 for 40-watt, and constant 24 for 60-watt lamps. 



74 ELECTRICAL AND MAGNETIC CALCULATIONS. 



And if 50 volt lamps are used 

</ 2 = 0.412 x Dn X 



100 



% 



% 



(37) 



Furthermore, if 



100 



% 



be worked out for the per 



cents drop ordinarily met in practice, and its value com- 
bined with the constant just found, the formula may thus 
be very much simplified. Thus suppose a drop of 5 per 



cent. 



_ 100 - o/o 
Then —. — '— =s 19, 



This multiplied by 0.098 
7o 
gives 1.86 as the constant for no volt 220 ohm lamps at 

5 per cent drop. The only variables are now D and n. 

For 50 volt lamps the total constant will be 0.412 X 19 

= 7.828 at 5 per cent; and (37) will be, at 5 per cent, 

d 2 = 7.828 X Dn. 

For no volt lamps there are set down the complete sim- 
plified equations for various per cents up to 10 per cent. 

For 1 % drop, d 2 = 9.7 X Dn ^ 

For \\of drop, d 2 = 6.44 X Dn 

For 2 % drop, d 2 = 4.8 X Dn 

For 2^% drop, d 2 = 3.82 X Dn 

For 3 °/ drop, d 2 = 3.17 X Dn 

For 7>\°lo drop, d 2 = 2.70 X Dn 

For 4 % drop, d 2 = 2.35 X Dn 

For 5 % drop, d 2 = 1.86 X Dn 

For 6 % drop, d 2 = 1.54 X Dn 

For 7 % drop, d 2 = 1.30 X Dn 

For 8 % drop, ^/ 2 = 1.13 X Dn 

For 9 % drop, ^/ 2 = 0.99 X Dn 

For 10 % drop, d 2 = 0.88 X Dn ) 



r 



(38)* 



* For 30-watt tungsten lamps multiply the constants in these equations by 0.55 
for 40-watt, multiply by 0.7s; for 60-watt, multiply by 1.1. 



WIRING FOR LIGHT AND POWER. 75 

Example. — Calculate the size of the leads for 200 
lamps at 300 feet for 4 per cent drop. Lamps always 
no volts unless otherwise specified. 

Solution. — d 2 = 2.35 x Dn = 2.35 x 300 x 200 
= 141,000 cir. mils = No. 00, A.W.G. 

The drop should be made as small as it can be con- 
sistent with economy in the cost of conductors ; (1) be- 
cause large drops are detrimental to good regulation ; (2) 
because large drops mean the same percentage waste of 
the energy of the generator. The former is the more 
important from the consumer's standpoint, while the latter 
reason appeals the more directly and strongly to the pro- 
ducer. 

Example. — Suppose a circuit has 100 lamps and the 
drop on mains and leads is 1 5 volts. What will be the 
voltage of the rest of the lamps when 5 o are turned out ? 
Determine the same for 5 volts drop. 

Solution. — Assume the lamps to have no volts pres- 
sure, average. Drop E = IR\ when 50 are cut out /is 
only half as large as before ; hence, since R is constant, 
E the drop will be only half as much as before, or 7^ 
volts. Therefore, until the machines are regulated to 
suit the new conditions the remaining fifty lamps will run 
at no + i\ = 117^ volts, which is excessive. Likewise 
any small change in the number of lamps will be accom- 
panied by a proportional variation of voltage in the rest. 

If the total drop at full load be only 5%, then the other 
50 lamps will run at 110 + \ of 5 = 112^ volts when one- 
half the load is taken off. 



j6 ELECTRICAL AND MAGNETIC CALCULATIONS. 

It is apparent then that uniformity of light under vary- 
ing load can best be secured by having the full-load line 
drop as small as can be made consistent with economy 
in the cost of wire. 

From this it should also be clear that the drop should 
be divided so as to decrease from the generator out 
through mains, leads, and lamp circuits. 

Wiring Diagrams. — It will often be convenient to consult 
the wiring diagrams for installing the line for no and 55 
volt lamps, Tables 3 and 4, instead of applying the formula 
in calculation. In these there are three horizontal and 
three vertical columns. The horizontal ones mean " lamp 
feet." The vertical ones mean " circular mils," except the 
underscored numbers, which mean the gauge number, 
B. & S., or A. W. G. The numbers in both sets of columns 
are so many thousands. In consulting the diagrams 
always use corresponding columns, inside-inside, middle- 
middle, outside-outside. 

Example. — Use the diagram, and determine the size 
of wire for 150 lamps, no volts, at a distance of 200 feet, 
drop 5%. 

Solution. — Dn = 200 x 150 = 30,000 lamp feet. 
Find 30 in the inside horizontal column, follow up vertical 
line to intersection of 5 per cent radial line, then move 
left to inside vertical column, and obtain 54,000 cir. mils, 
which is between No. 3 and No. 2, nearer No. 3. 

If the same number of lamps is to be wired for 55 
volts at the same drop, the above process on the 55 volt 
diagram obtains 222,000 cir, mils, or about 2 No. o wires 









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WIRING FOR LIGHT AND POWER. 



77 



in parallel, thus showing the poor economy of wiring 55 
volt lamps at a low per cent drop. 

If the drop be made 10 per cent for 55 volt lamps the 
wire will be No. o. 

In case the lamp feet are too large to be found in one 
of the horizontal columns, divide by 10, find the size of 
wire for this number of lamp feet, then multiply the circu- 
lar mils so found by 10. 

*Example. — Find the circular mils for a main to sup- 
ply 500 lamps at 2000 feet, at a loss of 10%. 

Solution. — Dn = 2000 x 500 = 1,000,000. One-tenth 
of this is 100,000. Find the circular mils corresponding 
to be 88,000. Multiply this by 10, making d 2 = 880,000 
cir. mils. This is equivalent to 8 No. o wires in parallel. 




27. The Three -Wire System. — By the three-wire system 
is meant one in which two dynamos are connected in 
series, one terminal of each being connected respectively 
to one of the wires of the circuit, while the common junc- 
tion of the machines also connects with a third or middle 

* In applying the no-volt diagram to tungsten lamps, find the circular mils as 
for carbon lamps, then use the multipliers suggested on page 74 for tungstens. 











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78 ELECTRICAL AND MAGNETIC CALCULATIONS. 

wire of the circuit. The system therefore requires three 
wires for its operation. The voltage between the outside 
wires is twice the voltage of a single machine, while that 
between the outside wires and the middle, or neutral wire, 
is equal to that of each machine. When the system is in 
operation two lamps are in series between the outside 
wires, thus requiring one-half the current which the same 
number of lamps would require on the two-wire system. 
The diagram, Fig. 5, will make the details of the circuits 
clear. 

The three wires are usually continued out to the lamp 
circuits, and these also often have three wires where there 
is a very large number of lamps on each circuit. The 
economy in wire is apparent from the following considera- 
tions. 

The voltage between the outside wires is double that in 

the two-wire system and, as has been shown, the current is 

one-half. Hence for a given per cent drop the volts drop 

will be double that for two wires. The resistance of the 

■p 
wire is the volts drop divided by the current, or i? = — 3 , 

where the subscripts denote the kind of system. But 

£ z = 2,E 2 ,and/ 3 = i.4 Hence R z = -^ = 4 -7^ That 

2 A ■* 2 

is to say, the resistance for the three-wire system is 4 times 
the resistance for the two-wire system. Therefore the 
cross section of each wire is \ that of each wire in the two- 
wire system. 

To determine the wire for this system, then, apply the 
rules for the two-wire circuit, and take \ the circular mils 
so found. 



WIRING FOR LIGHT AND POWER. 79 

Example. — Find the size of wire for 200 lamps, no 
volts, at 200 feet, for 5 per cent drop. 

Solution. — From (38), d 2 = 1.86 X Dn. 

Therefore 

d 2 = 1.86 X 200 X 200 = 74>4°° c i r - mils. 
\ of 74,400 = 18,600 cir. mils for 3 wires, No. 8, nearest. 

The middle wire could usually be made smaller ; but 
should the number of lamps on the two sides of the circuit 
be unequal, it has to carry the excess back to the machine. 
It is therefore nearly always made the same size as the out- 
side wires. The total copper then in the three-wire system 
is f as much as for the same number of lamps on the two- 
wire system ; for each lead in the former is \ of each in the 
latter, and 3 leads make f of each, or | of both leads in 
the latter. That is, if a certain system could be installed 
using 2 wires for $500.00, it could be put in with 3 wires 
for § of $500.00 = $187.50. This is partly offset, how- 
ever, by the extra cost of installing, and by the necessity 
of using two machines. 

Several different methods * have been devised to over- 
come the disadvantage of using two machines, though 
they are but little used, if at all. An armature may have 
two independent windings and two commutators to which 
the circuits may then be connected, as with two machines. 
In the second method, the generator is wound to give 220 
or 440 volts, and a storage battery is connected across 
the wires ; the neutral wire is then joined to the center of 
the battery. Again, a third brush may be set for the 
neutral wire to give half the potential of the machine. 

* See Crocker's Electric Lighting, Vol. II., Chap. IV., for complete descrip- 
tion and diagrams of the various three-wire systems. 



8o ELECTRICAL AND MAGNETIC CALCULATIONS. 

28. Original Problems. — 1. By the fall of potential 
method find the gauge number of the wire to be used as 
leads from house switchboard to lamp circuit junction box 
65 feet, to supply 200 lamps, no volts, at 2^ volts drop. 

Wire No. 3 B. & S. 

2. Suppose the lamps in the above problem were 50 
volt lamps wired for the same drop ; what size of leads 
must be installed ? Wire No. o B. & S. 

3. Determine the wire for the following system. From 
the house switchboard a set of leads is run north 100 
feet where a branch is taken off to a distributing point for 
20 lights; 75 feet farther north another branch is run off 
from leads for 15 lights; 50 feet still farther north on 
leads is a junction box for 10 lights. The drop is 2 \ 
volts to first branch, 1^ between the first and second, and 
1 volt between second and third. Lamps are no volts. 

Wire, first section, No. 7. 
Wire, second section, No. 9. 
Wire, third section, No. 13. 

4. In the same way determine the size of the lamp cir- 
cuit wires for a drop of 1 volt, if the lamps from the first 
junction are at an average distance of 50 feet, from the 
second 30 feet, and from the third 20 feet. 

First No. 10B.&S. 

Second No. 14 B. & S. ) Nothing less than 

Third No. 14 B. & S. ) 14 to be used. 

5. Use the formula for wiring, and calculate the mains 



WIRING FOR LIGHT AND POWER. 8 1 

for 500 incandescent lamps, no volts, at a distance of 
500 feet for a drop of 8% of the generator volts. 

Wire 2 No. 00 in parallel. 

6. The circuits for problem 5 run out from the distrib- 
uting switchboard as follows: one circuit north 100 feet 
for 100 lamps; one south 50 feet for 200 lamps; one 
west 100 feet for 80 lamps; one east 150 feet for 80 
lamps. Drop on leads 2^%. Determine the gauge number. 

North No. 4. 
South No. 4. 
West No. 5. 
East No. 3. 

Note. — No. 4 wire could be used without any appre- 
ciable disturbance of the regulation, and in practice 
probably would be used in each branch. 

7. Determine the lamp circuits in the last problem, 
drop 1^%, and draw complete diagram of the system, the 
following data being given. The north branch supplies 
from the junction point 10 circuits averaging 10 lamps 
each, distances from junction averaging 40 feet. The 
south branch connects with four junction boxes near 
together; from No. 1, 60 lamps are fed on 5 circuits 
averaging 50 feet long; No. 2, 30 lamps on 4 circuits 
averaging 25 feet ; No. 3, 80 lamps on 8 circuits, 30 feet ; 
No. 4, 30 lamps on 3 circuits, 40 feet. The west branch 
feeds two junctions; No. 1, 50 lamps on 4 circuits, 60 
feet; No. 2, 30 lamps on 5 circuits, 50 feet. The east 
branch supplies two junction boxes ; No. 1, 70 lamps on 
7 circuits, 40 feet; No. 2, 50 lamps on 4 circuits, 50 feet. 



82 ELECTRICAL AND MAGNETIC CALCULATIONS. 

North No. 1 6, use No. 14. 
No. 1, wire No. 14. 

^ th ^ ^°* 2 ' w * re "^°* I ^' use I4# 
No. 3, wire No. 17, use 14. 

^ No. 4, wire No. 14. 

,, 7 ( No. 1, wire No. 14. 
West < ' ^ 

( No. 2, wire No. 17, use 14. 

C No. 1, wire No. 16, use 14. 

East ^ XT -at 

^ No. 2, wire No. 14. 

8. Use the wiring diagrams and verify the results 
obtained in problems 1 and 2, for 30-watt tungstens. 

9. By means of the diagrams verify problems 3 and 4. 

10. Verify the results in problems 5 and 6 by means of 
the diagrams. Also calculate for 40-watt tungstens. 

n. Work out problem 7 by use of the wiring diagrams. 

12. A town is to be wired, 3-wire system. Determine 
the sizes of wire to be used ; also the machine volts and 
line drops. Street mains run from the station i mile 
north on Main street to the intersection of Union street, 
and supplies east and west branches ; the one east runs 
on Union street 500 ft., thence north on College street 
250 ft. ; the one west runs on Union street 500 ft, thence 
north on Congress street 250 ft. The Main street line 
also continues north 500 ft. to Washington street, whence 
it sends one branch east 500 ft., thence north on College 
street 250 ft. ; also one branch west 500 ft, thence north 
on Congress street 250 ft. The Main street line also 
continues north from Washington street 400 ft. to State 
street, whence one branch runs east on State street 500 



WIRING FOR LIGHT AND POWER. 83 

ft., and one west on State street 500 ft. There will be 
about 500 lamps, no volts, fed from the first junction on 
Main street, 300 on the east branch and 200 on the west 
branch. There will be probably 200 on the east branch 
at Washington street and 200 on the west branch. At 
State street there will be about 100 lamps fed east and 
100 west. It is decided to lose 8% on the Main street 
feeder mains and 4% in the cross street lines, leaving 
3 °f thence to lamps. See answers for °/ 's used. 

First section main feeder requires 5 No. 0000 

wires in parallel, 6°/ c drop. 
Second section 3 No. 000 wires in parallel, \\°f . 
Third section 2 No. 00 wires in parallel, f %. 
First east branch No. 00 wire. First west branch 

No. 1 wire. 
Second east branch and west branch No. 1 wire. 
Third east branch and west bra?ich No. 4 wire. 
Machine volts 130 X 2 = 260. 
Volts at branches 119.6. 
Volts at junc. lamp leads n 4.4. 
Volts at beginning lamp circuits n 1.8. 
Drop on mains 10.4 volts. 
Drop on branches 5 volts. 

13. Continue problem 12 as follows, drawing complete 
diagram. Each one of the branches described above 
feeds into the center of a north and south lead wire run- 
ning each way 125 feet to the center of lamp circuits. 
The first lead on College street supplies 150 lamps each 
side of feeding point, drop 2%. The second on Col- 
lege street supplies 75 lamps on the north and 125 on 



84 ELECTRICAL AND MAGNETIC CALCULATIONS. 

the south of feeding point. The third supplies 75 lamps 
south and 25 north of feeding junction. The first on 
Congress street supplies 150 lamps north and 50 south of 
feeding point. The second on the same street feeds 100 
lamps each way. The third, 40 lamps north and 60 south. 
Drop in all leads is 2%. 

First, College street, No. 6 wire each way. 

Second, College street, No. 7 S. and No. 10 N. 

Third, College street, No. 10 S. and No. 14 N. 

First, Congress street, No. 6 N. and No. 12 S. 

Second, Congress street, No. 8 N. and S. 

Third, Congress street, No. 12 N. and No. 10 S. 

14. There are installed at a distance of 300 feet from 
the street mains 5 motors in a factory: 2 of 20 H.P. 
each, 1 of 40 H.P. and 2 of 50 H.P. each. What size 
of wire must be installed for 5 °f drop, the motors being 
220 volt machines, and assuming the load not to exceed 
§ of the rated power ? 

2 No. 00 wires in parallel. 

15. A motor of 200 H.P. is employed at full load to 
operate a line of counter shafting in a foundry. A 
special wire is erected, running from the general power 
station 500 feet distant. The E.M.F. at the station 
switchboard is 250 volts, and the motor requires 220 
volts. Determine the size of wire to be installed. 

2 No. 00 wires in parallel. 



BATTERIES. 85 



VI. 

BATTERIES. 

29. Connection of Cells for Combined Output. — Applying 
Ohm's law to the electric cell, the current delivered is 

in which E is the electromotive force in volts measured 
between the electrodes when there is no circuit connected. 
R is the external, and r the internal resistance. E and r 
are called the " cell constants." 

Example. — Through what external resistance must a 
cell be connected whose constants are E = 1.5 volts and 
r = \ ohm, so that 1 ampere may be obtained ? 

Solution. — 



R + r R + 0.2 

Hence R + 0.2 = 1.5. 

Whence R = 1.5 — 0.2 = 1.3 ohms. 

Example. — How much current will flow in a circuit of 
5 ohms external resistance in which is placed a cell whose 
E.M.F. is 2 volts and internal resistance ^ ohm ? 

I =j£r r = : j^ = °-3 6 3 am P ere - 

Series Connection. — Whenever any number of E.M.F/s 
are connected together in series, the total E.M.F. is their 



86 ELECTRICAL AND MAGNETIC CALCULATIONS. 



sum if all are positive in the same direction ; or if some 
are reversed with respect to the rest, it is the difference 
between the sum of the positive and the sum of the nega- 
tive E.M.F.'s. The total resistance is the sum of all the 
resistances in series. Therefore if any number of similar 
cells n be joined in series, the current flowing through any 
external resistance R will be 

Example. — Draw a diagram of connections for 4 cells 
in series through an external resistance of 30 ohms, and 
calculate the current flowing when the cell constants are 
1.8 volts and 1 ohm. 



h - + - +. - + ■ ■ 



E-1.8 



r-i 



A/WWWWWWV 



/ = 



nE 



R-30 
Fig. 6. 

4 X 1.8 



R + nr 30 + 4 X 1 
For only one cell, 



= 0.212 ampere, 



1 = 



1.8 



0.058 ampere, 



3° + 1 
or a little more than 1 as much as 4 cells in series. 

Multiple Connection. — When any number of equal 
E.M.F.'s are placed in multiple arc, or parallel, the total 
E.M.F. is that of a single one, the effect of such an 



BA TTERIES. 



87 



arrangement being merely to reduce the internal resist- 
ance according to the law expressed in Chapter III. 
Therefore when any number of cells m are connected in 
parallel, the current flowing in an external resistance R 

will be E , N 

~ ' (4o) 



I 



R + 



m 



Example. — Draw a diagram of connections for 4 cells 
in parallel through an external resistance of 30 ohms, 
and calculate the current furnished by the battery thus 
formed when the cell constants are 1.8 volts and 1 ohm. 





wwwwvwvw 



/ = 



E 



R + 



m 



Fig. 7- 
1.8 

3° + - 



= 0.059 ampere, 



This is only an insignificant advantage over a single cell. 
Suppose the external resistance had been 1 ohm. The 
current would then be 

1.8 



/ = 



1 + 0.25 



= 1.44 amperes. 



This is about 1^ times the current furnished by a single 
cell through the same resistance. 



Example. — Suppose now the cells be placed in series 



88 ELECTRICAL AND MAGNETIC CALCULATIONS. 

through i ohm external resistance ; what current will be 
obtained ? 

r 4X1.8 

Solution. — I = = 1.44 amperes. 

1+4X1 ^ F 

This is exactly what was given when the cells were all in 

parallel. 

Example. — Use the cells first in series, then in par- 
allel upon 1 ohm external resistance. 

Solution. — For series, 

t 4 X 1.8 

I = = 1.60 amperes. 

0.25 + 4 

1 8 
For parallel, /"= -j- = 3.6 amperes. 

\j ' \j 

The parallel arrangement gives more than twice the 
current given by the series connection. 

Comparing the results of this problem with the pre- 
vious problem having 30 ohms external resistance, it is 
apparent that for large external resistances relative to the 
battery resistance, the series method gives the larger 
current ; while for relatively small external resistances, 
the parallel method may give the better results. 

Multiple-Series Connection. — Sometimes, however, a better 
advantage is obtained by combining these two methods of 
connection in what is called the multiple-series method. 
The principles are the same as before ; calling n the 
number in series and m the number in parallel, the 
formula for / becomes 

T _ U ^ 

R nr (4 1 ) 

m 



BA TTERIES. 89 

Example. — What current will be obtained through 
30 ohms external resistance from 4 cells connected 2 in 
series and 2 in parallel, cell constants being 1.8 volts and 
1 ohm ? Draw diagram of connections. 

nE 2 X 1.8 

Solution. — I = = = o. 1 1 6 amp. 




vwwwww 



R=30 
Fig. S. 

This current is about twice as much as when all were 
placed in parallel. 

Example. — Suppose the external resistance had been 
1 ohm, how much current would flow through it ? 

,.2 + 1.8 

Solution. — I = = 1.8 amperes. 

2x1 

1 H 

2 

This gives a larger current than either of the other 
methods of connecting the cells through the same re- 
sistance ; both the series and the parallel gave 1.44 
amperes. 

The Best Arrangement. — The question naturally sug- 
gests itself, which of these methods for a given number 
of cells will give the most current through a given 
external resistance ? As already suggested the matter is 



9<D ELECTRICAL AND MAGNETIC CALCULATIONS. 

roughly decided by a comparison of external and internal 
resistances. An observation of the numerical values 
given by the formulae for these three cases, when the 
external resistance was i ohm, shows that in the series 
connection the internal resistance, 4 ohms, is larger than 
the external ; in the parallel connection the internal, 
0.25 ohm, is smaller than the external ; in the parallel- 
series connection the internal is equal to the external. 
Hence the conclusion is that the maximum current is 
obtained from a given number of cells connected to a given 
resistance when they are so arranged that the internal resis- 
tance is equal to the external. 

This is shown also from the following considerations. 
Since nE E 



X + ? *+ r - 
m n m 

R r 

it is apparent that I will be a maximum when 1 

n m 

is a minimum, for the numerator E is constant, being 

R r 

the E.M.F. of one cell. Now the product of— and — = 

n m 

Rr 

— is a constant, R and r both being constant, and mn 
mn 

the whole number of cells in the battery. It is a well- 
known principle and can be easily illustrated by a sim- 
ple numerical example, that the sum of two quantities is 
the least, their product being a constant, when the two 

R r 

quantities are equal. Therefore 1 will be a minimum 

z n m 

R r nr 

when — = — ; that is when mR = nr, and R = But 

n m m 



BATTERIES. 9 1 



nr 



is the internal resistance of the battery. Therefore 
m 

the external and i?itemaZ resistances must be equal for the 
greatest rate of flow. But under these circumstances only 
one-half the total energy is expended in the external cir- 
cuit, the other half being spent in the battery resistance. 
The efficiency of a battery arranged for greatest current 
is therefore 50%. 

30. The Best Arrangement for a Required Current. — 
To find the number of cells for a required current it is 
only necessary to apply the general formula for the con- 
nection of cells, 11E 



„ nr 

R+ — 

m 



remembering that for the greatest current the cells must 

nr 

be arranged so that R = 

m 

Example. — How many cells arranged in series are 
necessary for 4* amperes through a resistance of \ ohm, 
the cell constants being 2 volts and \ ohm ? 

Solution. — Taking n = the number of cells in series 

and m = the number in parallel, we have 

11E 2 n , N 

/= = = 4| (1). 

nr 1 n b K J 

R + — - + - 
m 2 4 

m = i, all being in series. Clearing (1) 

2 n = 2.4 + i-2 n\ 

and 0.8 n = 2.4 ; n = 3 cells. 

Example. — How many cells in parallel will give 
3.6 amperes through an external resistance of J ohm, 
cell constants 1.8 volts and 1 ohm? 



92 ELECTRICAL AND MAGNETIC CALCULATIONS. 

c _ nE I X 1.8 

Solution. — 7 = = = 3.6 amperes. 

#24 ;/z 

From this 1.8 m — 0.9 #2 = 3.6 ; and m = 4 cells. 

Example. — How many cells and what arrangement 
will be necessary to supply 3 amperes, the external resist- 
ance being if ohms, and the cell constants 2 volts and 
1 ohm? 

o T nE 2n N 

Solution. — /= = = 3 (1), 

^ nr n 

E + — if + ~ 
m m 

and 2 /z = c + — (2). 



nr ^ n . 

Since — = R = if, #z = — - , since r = v 
w 3 if 



Substituting this value of ^ in (2), 

3 n 

2/ * = S + V = S + 5 = IO ' 



and ^ = 5 ce ^ s m series. 

Also m = —- = -^- = 5 x f = 3 cells in parallel. 



Hence it will require 15 cells, 5 series, 3 parallel. 

For a required current at a given voltage the following 
rules may be applied with satisfaction. 

Divide the cell E.M.F. by its resistance, (a) If this short 
circuit current is twice or more than twice the required cur- 
rent, the cells will all be in series, and the number of cells 
will be E 1 , N 



BA TTERIES. 93 

E x is the required external E.M.F., E is the cell E.M.F., 
r the internal resistance, and / is the required current. 

(b) If the short circuit current is less than twice the re- 
quired curre7it and more than its equals or less than equal 
and more than its half, or less than the half a?id more thaii 
its fourth, and so on, place as many cells in parallel as 7vill 
make the short circuit current twice or more than twice the 
required current, then apply (42) for the number to be used 
in series, 

(c) If the short circuit cu?'re?it is 1, \, \, \, etc., times 
the required curre?it, place as many cells i?i series as will 
make the E.M.F. double the required line voltage, and the?i 
put enough in parallel to make the battery resistance equal to 
the external resistance, — Sloane.* 

Example. — There are 4 50-volt, 50-ohm lamps in par- 
allel. Cell constants are 1 .8 volts average and I ohm. How 
many cells and what arrangement will supply the lamps ? 

1 8 
Solution. — Short circuit current = -j- =9 amperes. 

5 

Required current /= f§ X 4 = 4 amperes. 

Hence, using (a), 

£1 5° 



n = 



-- = 50 cells in series. 



E — Ir 1.8-4X5 

Example. — Using similar cells, how many will it take 

for 8 lamps similar to the above ? 

• • 1.8 

Solution. — Short circuit current = -=- = 9 amperes. 

5 

Required # current / = %% X 8 = 8 amperes. 

* Sloane, Arithmetic of Electricity , p. 69 et seq. 



94 ELECTRICAL AND MAGNETIC CALCULATIONS. 

Hence, using (<£), place 2 cells in parallel giving 

9 X 2 = 18 amperes, cell current. 
Whence 

E \ 5° 11 ■ 

n = — — = —t £ — z r = ko cells in series. 

E-Ir i.8-8x|x| 

The whole number required is mn = 2 x 50= 100 cells 
arranged 2 in parallel and 50 in series. 

Example. — Suppose there were 18 lamps in parallel, 
how many cells would be necessary ? 

Solution. — Short circuit current = 9 amperes. 
Required current /== |-g. x 18 = 18 amperes. 
Hence, using (V), we must place in series 

n = 2 -g± = ^-p = 5 6 cells > 

giving an internal resistance of 

-5- X 56 = ni ohms. 

Therefore we must place in parallel cells enough to re- 
duce to AO. = 2-I ohms. Hence m in parallel equals 
ni-f-2-I = 4 cells. Therefore there will be required 
mn = 4 x 56 = 224 cells. 

31. Arrangement of Cells for a Required Efficiency. — 

As for any generator of electrical energy, the efficiency of a 
battery is the ratio of the external energy to the total energy 
developed, and this is the same as the ratio of the external 
to the total resistance in the circuit. 



BA TTERIES. 95 

Eff. = -tA- ' Ui) 

R+r K * 6J 

In order to obtain a formula for r, multiply (43) by R + r, 
when R x eff. + r x eff . = R ; or 

*(,-eff.) . 
r - dl (44) 

Example. — What is the efficiency of a battery deliv- 
ering current through an external resistance of 50 ohms 
when the battery resistance is 10 ohms ? 

Solution. — Eff. = —^— = — ^ — = 831%. 

R + r 50+10 ° 3 ' 

Example. — What must be the internal resistance of a 
battery when the external is 60 ohms in order that the 
efficiency will be 66f% ? 

iP(i-eff.) 60(1 -.66 J) 

Solution. — r= — - — = — - — — — — — = xo ohms. 

eff. o.66§ ° 

R (1 - eff.) . . . 

Since r = — - — = gives the internal resistance, the 

en. 

total resistance is expressed evidently by 

™ , n H(i — eff.) 

Total resistance == R H — -= '- ■ 

en. 

If the required current is /, then the E.M.F. of the battery 

to give a certain current through a known resistance at a 

required efficiency will be, from Ohm's law, 

R (1 - eff.)" 



- 



E = \S + 



eff. 



/• (45) 



Example. — Required to provide storage cells for 450- 
volt, 50-ohm incandescent lamps in parallel. Cell con- 
stants 2 volts and T \j- ohm, the battery to work at 75% 
efficiency. 



96 ELECTRICAL AND MAGNETIC CALCULATIONS. 
Solution. — 

Hence the number of cells in series will be 

66.6 



n = 



= 34 cells. 



Also for 75% efficiency the cells must be arranged so 

that 

i?(i-eff.) ^(1-0.75) , . 

r = iv = ^-^ — = 4.16 ohms. 

eff. 0.75 

But the internal resistance of the 34 cells in series is 
34 x ro = 3-4 onms > or already less than is required for 
75%. Therefore 34 cells in series meet the require- 
ments and the efficiency will be 

Eff. = I2 j_ 5 =78%. 
12.5+3.4 

32. Charging Storage Cells. — In charging batteries 
the E.M.F. applied must be equal to the ohmic drop in 
the wires and cells, plus the battery E.M.F. which acts as 
a counter electromotive force ; it also must vary with the 
condition of charge in the cells, rising as the battery ap- 
proaches full charge. 

Example. — A battery of 45 cells is arranged in series 
for charging. Each has an internal resistance of 0.005 
ohm and a counter E.M.F. of 2.1 volts. The leads have 
a resistance of 0.05 ohm. The cells take a charging cur- 
rent of 100 amperes ; what must be the terminal voltage 
of the dynamo ? 



BA TTERIES. 97 

Solution. — 

The ohmic drop = (0.005 x 45 + °-°S) X 100 

= 27.5 volts. 
Counter E.M.F. = 2.1 X 45 = 94-5 volts. 

Hence the total voltage applied from dynamo switch- 
board must be 

E = 94.5 + 27.5 = 122 volts. 

Example. — Find the current at the beginning and end 
of the charge of 40 cells in series, each having 0.002 ohm 
internal resistance, leads 0.07 ohm, when the charge lasts 
6 hours and the cell E.M.F. changes during charge from 
2 volts to 2.5 volts. The generator remains uniformly at 
125 volts. 

Solution. — 

Counter E.M.F. at first = 40 x 2 = 80 volts. 
Counter E.M.F. at end = 40 x 2.5 =100 volts. 
Resistance of circuit = (40 x 0.002) + 0.07 

= 0.15 ohm. 
Ohmic drop at first = 125 — 80 = 45 volts. 
Ohmic drop at end =125 — 100 = 25 volts. 

Hence from Ohm's law, 

T £ 1 45 

J 1 = —- = —— = 300 amperes, 
K 0.15 

and _ £ 2 25 

Jo = —pr = = 166.66 amperes. 

R 0.15 

33. The E.M.F. of Cells from the Available Heat of 
Chemical Action. — The source of the electrical energy of 
a cell is the chemical energy transformed in it. Neglect- 



98 ELECTRICAL AND MAGNETIC CALCULATIONS. 

ing the losses which occur more or less in all cells, the 
electrical energy obtained is equivalent to the energy of 
chemical action. Now when elements combine or sepa- 
rate heat is liberated or absorbed. Therefore we have a 
measure of the chemical energy in the amount of heat ex- 
changed. The amount of an element separated is also 
strictly proportional to the quantity of electricity flowing. 
These considerations furnish the principles for determining 
the E.M.F. necessary to separate certain chemical com- 
pounds, or produced by known chemical separations and 
combinations taking place in a cell. 

The atomic weight of an element is the weight of its 
atom relative to that of hydrogen, H, which is i. 

The valence of an element is the number of hydrogen 
atoms to which it is equivalent in combining power, or the 
number of hydrogen atoms which would directly combine 
with or replace it. Thus water, H 2 0, consists of two 
atoms of hydrogen held to one atom of oxygen. Hence 
O has a valence, or holding power, of 2. 

The chemical equivalent of an element is its weight rela- 
tive to hydrogen which enters into combination. In other 
words, it is the weight of the element which would com- 
bine with unit weight of hydrogen. It is obtained by 
dividing the atomic weight by the valence. 

For example, in H 2 the atomic weight of O is 15.96 ; 
that of H is 1. But 15.96 of O is equivalent to 2 x 1 = 2 
of hydrogen. Hence each unit of H is equivalent to 
15.96-^2 = 7.98 of O ; that is to say, each gram of H will 
require for combination 7.98 grams of O. The chemi- 
cal equivalent of oxygen is 7.98, or its atomic weight 15.96 
divided by its valence 2. 



BA TTERIES. 99 

The electrochemical equivalent of an element is the 
weight of it expressed as the fraction of a gram which will 
be separated by one coulomb of electricity. Clearly the 
electrochemical equivalent is proportional to the chemical 
equivalent. Thus i coulomb separates 0.00001038 gram 
of hydrogen, also 0.0000828 gram of oxygen, and these 
are in the same ratio as 1 to 7.98, or as the chemical 
equivalents. If we take the reciprocal of the electro- 
chemical equivalent we obtain the quantity of electricity 
necessary to separate 1 gram of the element. See table 
of chemical constants. 

Example. — Obtain the formula to be used in obtaining 
the electromotive force in volts produced in any cell whose 
thermal equivalents are known ; that is when the calories 
of heat formed or absorbed in the chemical reactions can 
be obtained. 

Solution. — The electrical energy developed in any 

cell is 

Energy = Eli. 

The available heat energy for transformation into electri- 
cal energy is 

Heat=/.#; 

in which H is the free heat in calories due to chemical 
action, and ./is Joule's Equivalent, or the ergs of energy 
represented by each unit of heat = 4.19 x io 7 . If ex- 
pressed in ergs the electrical energy will be 

E e = Elt x io 7 ergs. 
Therefore 

JH= Elt x io 7 = 4.19 x io 7 x H. 



IOO ELECTRICAL AND MAGNETIC CALCULATIONS. 



5. Chemical and Electrochemical Constants. 









H 






1 Elec- 








X 





Electro- 


trochemi- 


Names. 




M 


w 
u 

w 




Chem. 
Equiv. 


chemical 

Equivalent, 

in Grams 


cal equi- 

VALENT, 

or Cou- 




a 


< 




H 




per Coulomb. 


lombs 




tfl 


> 


< 






per Gr. 


Electropositive EL 














Hydrogen . . 


H 


I 


1. 00 


I. OO 


O.OOOOIO38 


96,340.0 


Potassium . . 


K 


I 


39-03 


39.03 


O.OOO4051 


2,469.0 


Sodium . . . 


Na 


I 


23.00 


23.OO 


O.OO02387 


4,189.0 


Silver. . . . 


Ag 


I 


107.70 


IO7.70 


O.OOII18 


894.5 


Copper (ic) . . 


Cu 


2 


63.18 


3 J -59 


O.OOO3279 


3,050.0 


Mercury (ic) 


Hg 


2 


199.80 


99.90 


O.OOIO37 


964.3 


Tin (ic) . . . 


Sn 


4 


117.40 


2 9-35 


O.OOO3046 


3,283.0 


Iron (ic) . . . 


Fe 


3 


55.88 


18.63 


O.OOOI933 


5,171.0 


Nickel (ic) . . 


Ni 


2 


58.60 


29.30 


O.OOO3041 


3,287.0 


Lead .... 


Pb 


2 


206.40 


103.20 


O.OOIO7 1 


933-7 


Zinc .... 


Zn 


2 


64.88 


3 2 -44 


O.OOO3367 


2,970.0 


Electronegative El. 














Oxygen . . . 


O 


2 


15.96 


7.98 


O.OOO08283 


12,070.0 


Chlorine . . . 


CI 


1 


35-37 


35-37 


O.OOO367 


2,724.0 


Nitrogen . . 


N 


3 


14.01 


4.67 


O.OOOO4847 


20,630.0 



From which Elt = 4.19 H. 

But It, the number of coulombs corresponding to 1 gram 
of hydrogen, or 1 chemical equivalent of any other ele- 
ment, as 107.7 grams of silver, 31.59 grams of copper, etc., 
is 96,340 coulombs ; see table 5. Hence, since 

It— 9 6 >34-o, 
E x 9 6 >34o = 4.19^ 
or • E = 0.000043 H* (46) 

If H is expressed in kilogram- degree calories, instead of 
gram-degree calories, (46) becomes 

E = 0.043 H- (47) 

* Exactly, E — 0.000043 H -f- kT, where k is the temperature coefficient of 
E.M.F. of the cell, and T is the absolute temperature. 



BATTERIES. IOI 

We thus obtain the electromotive force from the result- 
ant heat of chemical action, keeping in mind that the 
result can only be approximate because of local action 
and heat absorbed from or given up to surroundings. 
The heats of combination of many ions, that is atoms or 
groups of atoms, have been 'determined. See table 6 for 
the heats resulting from the chemical reactions taking 
place in the most common cells. It is to be observed 
that H in the formulae is the algebraic heat ; that is, the 
difference between the heats of formation and those of 
disintegration in the cell. 

Example. — What is the minimum voltage a cell may 
possess to decompose water, H 2 ? 

Solution. — i gram H combining with 7.98 grams O 
liberates 36,500 calories; see table 6. Hence to separate 
them an equivalent amount of energy is necessary. 
Therefore 

E = 0.000043 x 36,500 = 1.56 volts. 

Example. — What is the approximate E.M.F. of the 
Daniell cell ? 

Solution. — In this cell the elements are 

Zinc I sulphuric acid copper sulphate | copper. 

The chemical reactions are 

Zn + H 2 S0 4 = ZnS0 4 + H 2 . (1) 

CuS0 4 + H 2 = H 2 S0 4 + Cu. (2) 

Calories. 

'Heat liberated by the formation of ZnS0 4 = 53,500 

(i)-j Heat absorbed by the separation of H 2 S0 4 = 36,5oo 

.Heat available for transformation =17,000 



102 ELECTRICAL AND MAGNETIC CALCULATIONS. 

Calories. 

'Heat liberated by the formation of H 2 S0 4 = 36,500 

(2)^ Heat absorbed by the separation of CuS0 4 = 28,400 

^Available for transformation = 8,100 

Total available heat = 17,000 + 8,100 = 25,100 calo- 
ries. 

Therefore .# = 0.000043 X 2 5> IO ° = I -°793 volts. 
64. Heats of Formation or Separation. 



Compounds. 


Ions. 


Calories 
Heat. 


H 2 S0 4 


H 2 and S0 4 


36,500 


ZnS0 4 


Zn and S0 4 


53'500 


H2O 


H 2 and O 


36,500 


HC1 


H and CI 


19,650 


ZnCl 2 


Zn and Cl 2 


56,400 


PbCl 2 


Pb and Cl 2 


39,200 


CuO 


Cu and O 


20,200 


Zn(OH) 2 


Zn and 2 (OH) 


41,800 


Zn(OH) 2 +2KOH 


Zn(0K)2 and 2 H 2 


8,000 


PbS0 4 


Pb and S0 4 


37400 


AgCl 


Ag and CI 


14,600 


PbO 


Pb and O 


25,500 


Pb0 2 


Pb and 2 


3^570 


co 2 


C and 2 


51,3°° 


CuS0 4 


Cu and S0 4 


28,400 



34. Material Consumed in a Battery. — It is sometimes 
interesting, if not essential, to compute the amount of 
chemicals used in a cell or a battery for a given amount 
of work done by the cell. For this purpose the following 
may be used with fair accuracy. 

The electrochemical equivalent in grams per coulomb 
multiplied by the number of molecules of the compound, 
or the number of atoms of the element going into combi- 
nation, will give the number of grams consumed per coulomb, 



BATTERIES. 103 

or per ampere-second. If this result be divided by the 
voltage of the cell, the result will be in grams per watt. 
This now multiplied by 3600 will give the number of grams 
used per watt-hour. 

Hence Z X N , , ox 

w = — — — X 3600. (48) 

In this w is the weight in grams per watt-hour ; Z is the 
electrochemical equivalent ; N is the number of mole- 
cules or atoms entering combination and is obtained 
from the chemical equation ; E is the cell E.M.F. 

Example. — How much zinc is used up in a Daniell 
cell working on a total resistance of 10 ohms for 1 hour ? 

Solution. — For zinc, Z = 0.0003367 ; E = 1.08 volts. 
/ = — — = 0.108 ampere. The chemical equation is 

Zn+ H 2 S0 4 = ZnS0 4 + H 2 . 

Hence there is one atom of Zn entering into combination. 

Hence 

0.0003367 X 1 . 

w = ^~i X 3600 = 1.12 grams per watt-hour. 

1.08 

Total watt-hours = 1.08 x 0.108 x i=E x ^X hrs. 
= 0.11664. 

Therefore total zinc dissolved in 1 hour is 

W = 1. 12 x o. 1 1664 = 0.1306 grams. 

The work may be slightly simplified by using a formula 
for total weight at once, obtained by multiplying the 



104 ELECTRICAL AND MAGNETIC CALCULATIONS. 

formula for grams per watt-hour by the formula for the 

number of watt-hours. 

ZxiV^X36oo __ . rri ^ TT 

W == ■=-£ x EI x hours = 3600 ZNI X hours, 

or W= 3600 ZNIT, ■ (49) 

in which W is the total weight separated or combined, N 
is the number of atoms or molecules combining, /is the 
current in amperes, and T is the time in hours during 

which the current flows. If E and R are known — may 

be put in place of /. The formula may be still further 
shortened by getting Z, the electrochemical equivalent, in 
grams per coulomb-hour, then (49) becomes 

W=ZNIT. (50) 

Example. — How much CuS0 4 is used up under the 
conditions named in the last problem ? 

Solution. — The chemical equation is 

CuS0 4 + H 2 = H 2 S0 4 -f- Cu. 

Hence 1 molecule of CuS0 4 is separated, and N== 1. 
For copper, Cu, at. wt. = 63.18. For (S0 4 ) at. wt. = 
32 + (15.98)4 = 95.92. Total molecular weight CuS0 4 = 
63.18 + 95.92 = 159. 1. Z for Cu = 1. 18 grams per 
coulomb-hour. For CuS0 4 , 

Z = 1. 1 8 X s = 2.9 grams per coulomb-hour. 
63.18 

Therefore from (50) 

W '= ZNIT= 2.9 x 1 X 0.108 x 1 = 0.313 gram. 



BATTERIES. 105 

35. Original Problems. — 1. What is the greatest ex- 
ternal resistance that may be used when the battery con- 
stants are 5 volts and 2 ohms, in order that \\ amperes 
may be obtained ? R — 1.33 ohms. 

2. It was observed that a cell whose electromotive 
force is 1.8 volts gave 1 ampere through an external re- 
sistance of 1 ohm; what was the internal resistance of the 
cell ? r = 0.8 ohm. 

3. What is the E.M.F. of a cell whose internal resis- 
tance is 1 ohm, when 1 ampere flows through an external 
resistance of 1 ohm ? E = 2 volts. 

4. The cell constants are 2 volts and \ ohm. Re- 
quired 5 amperes through an external resistance of 1.5 
ohms ; how many cells joined in series will be required ? 

n = 10 cells. 

5. Can 5 amperes be obtained through the same resis- 
tance with the cells joined in parallel ? Show why. 

6. What must be the internal resistance of each cell 
and of the whole battery when 5 cells joined in parallel 
through an external resistance of 0.835 onm gi yes 2 
amperes, E being 1.75 volts ? Each cell = ^ ohm. 

Battery = ^ ohm. 

7. Which is the better arrangement of 4 cells whose 
constants are two volts and \ ohm, series or parallel, 
when the external resistance is 20 ohms? Is there any 
better arrangement of 4 cells ? (a) Series, (b) No. 

8. What is the largest external resistance that can be 

used in 7, in order that the parallel connection will give 

a current equal to the series connection ? 

R = 0.5 ohm. 



I06 ELECTRICAL AND MAGNETIC CALCULATIONS. 

9. What current will be obtained under the conditions 
of problem 8 when two of the cells are placed in series 
and 2 in parallel? ^=4 amperes. 

10. Find the number of these cells required in parallel 
when 5 are placed in series to supply small lamps of 
5 ohms resistance with 1.8 amperes of current. 

m = 5 cells in parallel. 

11. Find the whole number of cells to be used under 
the following conditions : cell constants 2 volts and 
\ ohm external resistance 10 ohms; when n are put in 
series and m in parallel the current is 0.97 ampere ; but 
when m are put in series and n in parallel only 0.78 

ampere is obtained. mn = 4 x 5 = 20 cells. 

• 

12. Group the cells in problem 11 so as to derive the 
greatest possible current under the conditions given. 
Determine the efficiency in 11 and 12. 

Place 20 in series. 

Eff. 11 =97.5%. 
Eff. 12 = 71.4%. 

13. What is the least number of cells and their 
arrangement for 2 amperes through an external resis- 
tance of 1.5 ohms, cell constants 2 volts and \ ohm ? 

11 = 3 cells in series. 
m = 1 cell in parallel. 

14. Ninety cells whose internal resistance is each 
1 ohm are available for a circuit whose resistance is 
10 ohms ; arrange the cells for the greatest current. 

n = 30 cells in series. 
m = 3 cells in parallel. 



BATTERIES. \OJ 

15. What is the least number and the arrangement of 
cells of 2 volts 4 ohms each, when 1 ampere is desired in 
an external resistance of 15 ohms? 

n = 15 cells ; m = 4 cells. 

16. There are 4 uo-volt, 220-ohm lamps in parallel; 
storage cells are at hand of 2 volts and I ohm each. How 
many cells will it require to give the necessary current ? 

n = 69 cells in series. 

17. Suppose the cell constants were 2 volts and 1 ohm ; 
what number and arrangement would be required in 
problem 16? n = no ; m = 2 ; total 220. 

18. If the E.M.F. required in the circuit be 50 volts 
and the current 5 amperes, cell constants 2 volts and 
\ ohm, how many cells will be required ? 

n = 37 ; m = 2 ; total 74 cells. 

19. Solve problem 15 for 60% efficiency. 

n = 13 ; m = 5. 

20. How many cells would be required in problem 16 
to work at 75% efficiency ? n = 74 cells. 

21. Solve problem 17 for 80% efficiency. 

n = 69 ; m = 5 ; total 345. 

22. What is the least voltage a cell may have to be 
used for decomposing hydrochloric acid, HC1, when 
platinum electrodes are used ? E = 0.84 volt. 

23. Zinc and platinum are used as electrodes in simple 
HCL What will be the maximum E.M.F. of the cell? 

E = °-7353 vol t' 



108 ELECTRICAL AND MAGNETIC CALCULATIONS. 

24. Suppose copper sulphate, CuS0 4 be used in the 
last cell in a porous cup with a platinum, Pt, electrode. 
What will be the cell E.M.F. ? 

E = 1.0836 volt. 

25. The final action in the " silver chloride " cell is 
represented by the equation 

2 AgCl + Zn = ZnCl 2 + 2 Ag. 
What is its E.M.F. ? E = 1.17 volts. 

26. How much zinc, Zn, will be used in 10 hours 
when 10 Daniell cells are connected 5 in series and 2 
in parallel through a resistance of 5 ohms, cell constants 
being about 1.08 volts and 2 ohms? 

Zn used = 6.54 grams. 

27. How much CuS0 4 will be used in the above ar- 
rangement? CuS0 4 used = 15.66 grams. 

28. The final action in the Becquerel cell is represented 
by the equation, PbS0 4 + Zn = ZnS0 4 + Pb. 

Determine the E.M.F. and the amount of Pb separated in 
a cell working 10 hours on 1 ohm resistance? 

E = 0.69 volts. 

29. Find the E.M.F. of the Edison-Lelande cell whose 
chemical formulae are as follows : 

(1) Zn + 2H 2 O =Zn (OH) 2 + H 2 . 

(Zn (OH) 2 + 2KOH = Zn (OK) 2 -f 2 H 2 0. 
**' I H 2 + CuO = H 2 + Cu. 



BATTERIES. IO9 

Solution. — In the first and second reactions 41,800 + 
8000 calories are set free. In the last reaction 20,200 
calories are absorbed. Hence the available heat is 49,800 
— 20,200 = 29,600 calories. Hence 

E = 29600 x 0.000043 = I,2 7 v °lt- 

30. Twenty small storage cells are to be charged with 
a minimum of 10 amperes. If the highest cell E.M.F. is 
2.5 volts, cell resistance 0.008 ohm, and wire resistance 
0.14 ohm, with what voltage should the cells be charged ? 
If a no-volt incandescent machine is the only one avail- 
able, how can arrangements be made to charge the cells 
by means of it ? 

(a) Charging E.M.F. = 53 volts. 

(b) Place 10 50-volt lamps in parallel, and 

in series with the cells to be charged ; 
then reduce the machine voltage by 
means of its field rheostat till 10 am- 
peres are obtained. 

31. There are 50 storage cells whose constants may 
be taken at 2.1 volts and 0.002 ohm to be charged from 
a 250-volt circuit. The connecting wires, leads, etc., 
have about 0.9 ohm resistance ; how much additional 
resistance must be added in series so that the cells may 
be charged with about 40 amperes of current ? 

R = 2.6 ohms. 

32. How many storage cells may be charged in series 
on a no-volt circuit when 50 amperes are required, the 
cell E.M.F. being about 2 volts, resistance 0.004 ohm, 
leads and connecting wares 0.088 ohm ? 

Number in series = 48 cells. 



IIO ELECTRICAL AND MAGNETIC CALCULATIONS. 



VII. 
MAGNETISM. 

36. Units and Definitions. — ( a ) Magnetic flux is analo- 
gous to current flow in electrical units. In a magnetic 
field the flux originates at a north pole and passes 
through space into a south magnetic pole, thence through 
the magnet to the north pole, thus completing the mag- 
netic circuit. If we conceive a unit pole as previously 
defined to be placed at the center of a spherical surface 
whose radius is 1 centimeter, then the whole flux passing 
through this spherical surface will be 47r lines of force, 
or 47r maxwells ; for unit pole will produce unit intensity, 
or one line per square centimeter, at unit distance, and 
there are 47rr 2 units of surface in a sphere ; in this case 
47rr 2 = 47r square centimeters, hence \tt lines total flux. 

The total magnetic flux is represented by the Greek 
letter <£, and is expressed in maxwells. Thus a magnetic 
pole sends out 500,000 lines of force ; the flux is 

<j> = 500,000 maxwells. 

( b ) The density of magnetic flux, or the density of the 
magnetization, or simply, the magnetic induction is the 
number of lines per square centimeter or per square 
inch, and is represented by B ; it is expressed in gausses 
when the area is in square centimeters. For example, 
suppose the magnet carrying the 500,000 maxwells of flux 



MAGNETISM. Ill 

have ioo square centimeters across sectional area, then 
the intensity of magnetization is 



B _ 500,000 = 



100 



000 gausses. 



In general d> 

6 B = - A , or <£ = ^« 

( c ) The intensity of field, or strength of field, is the num- 
ber of gausses in air, and is denoted by the symbol H. 
Thus the flux in 100 square centimeters of air is <f> == 2000 
maxwells ; the intensity of field is JT = ^^>- = 20 gausses. 
The intensity of induction in a bar of iron placed in 
this field is B = ZT/x, in which fx is the permeability of 
the iron, or the ratio of its intensity to that of the field. 
Let the permeability of a certain piece of iron be 250. 
Then 

B = 20 X 250 = 5000 gausses. 

(d) The energy due to magnetization varies as the 
square of the field density, or as H 2 . This is analogous 
to mechanical energy which varies as the square of velo- 
city, V 2 . This relation only holds good for air free from 
any metallic influences. Take a field density of 200 
gausses; then the energy in each cubic centimeter of 

space is, from Maxwell,* 

H 2 
Energy= 8V ^ Sl) 

Putting in known quantities we get 

(200) 2 
Energy per c.c. = g ^ '^ = 1592 ergs. 

* Maxwell, A Treatise on Electricity a?td Magnetism, Vol. II. Art. 633, 



112 ELECTRICAL AND MAGNETIC CALCULATIONS. 

In 10 cc. the energy would be 

15,920 . , 

1592 X 10= 15,920 ergs = 7 = 0.001592 joule. 

(e) Magnetic reluctance is represented by (R, and the 
unit is the oersted. It corresponds exactly to electrical 
resistance, and is the force opposing the magnetic flux. 
The reluctivity, or specific reluctance, corresponds to specific 
resistance, and is the reluctance of 1 cc. of the material 
forming any part of the magnetic path, or circuit. Hence 
the total reluctance in oersteds is 

& = £-. (52) 

a vo J 

Here k is the reluctivity, / the length of the portion con- 
sidered, a the cross section. The reluctivity of vacuum 
is unity. This is practically true of air, wood, copper, 
glass, paper, and in fact all non-magnetic substances. 
For iron and other magnetic substances it is less than 1 
but varies with B. The reciprocal of reluctivity is perme- 
ability, and is expressed by ^, as previously explained. 

1 

(f) Magnetomotive force is analogous to electromotive 
force. It is the force setting up the magnetism or mag- 
netic flux, and is represented by M.M.F., or simply by M. 
It is expressed in gilberts. Suppose the flux required is 
(fy = 500,000 maxwells, and the magnetic reluctance of the 
circuit is 5 oersteds. The M.M.F. would then be if = 
<£(R = 500,000 X 5 = 25 X 1 o 5 gilberts. While the abso- 
lute unit is the gilbert, the practical value for electromag- 



MAGNETISM. 113 

netism is given in ci7npere-turns , or the number of amperes 
multiplied by the number of turns in the coil = 11I. But 
M — i.2$6nf, so that the absolute value is easily ob- 
tained from the ampere-turns. Gilberts = 1.256 X ai7ipere- 
turns. 

( g ) The magnetizing force in air is the intensity of the 
field H. When applied to a solenoid or electromagnet 
it means the M.M.F. per centimeter of length of the coil 
expressed in gilberts, and is still represented by H. Thus 
a coil of wire 10 centimeters long develops an M.M.F. of 
5000 gilberts. Here the magnetizing force is H — h -\%^- 
= 500 gilberts per centimeter; but since the reluctance 
of 1 c.c. of air is 1, 500 is also the flux per square centi- 
meter, or H. This is the M.M.F. which sets up the 
magnetism and maintains it in 1 centimeter of the coil's 
length. 

In general ff= M ^ 

( h ) By magnetic leakage is meant that portion of the 
flux which passes through such paths as to be unavailable 
for the purposes for which the electromagnet has been 
constructed. Thus in a dynamo the leakage lines are 
those passing around the armature, or across from leg to 
leg, so as not to be cut by the wires of the armature. If 
cj>j is the flux in the field cores, and cf> a is that passing 
through the armature core, the coefficient of leakage is 

In dynamos v varies from 1.1 to 1.7. Say for average 
machines, v = 1.35. 



114 ELECTRICAL AND MAGNETIC CALCULATIONS. 

( i ) A magnet is saturated when it has become so 
strongly magnetized, and its permeability is therefore so 
reduced that further magnetomotive force does not appre- 
ciably increase the flux. As the current is made to 
increase in the winding of an electromagnet the flux 
rapidly increases at first, its permeability decreasing as 
the intensity of magnetism increases ; the increase in flux 
becomes less rapid as the magnet approaches saturation, 
until finally further increase of current fails to produce 
any greater intensity. If now the current decrease, the 
magnetism falls though less rapidly than the current to 
a certain point below which it does not go even when the 
current becomes zero. The amount of magnetism thus 
remaining is called remanent or residual magnetism. 

( j ) Magnetic hysteresis is the frictional resistance to 
the turning around of the iron molecules which takes 
place during magnetization or reversals of magnetization. 
This explains why the change in the magnetism lags be- 
hind the change in the magnetizing current as stated 
above. Hysteresis is especially to be noted in iron sub- 
jected to rapidly alternating magnetizing forces, as in 
armatures of dynamos and in alternating current trans- 
formers. It varies with the frequency and with the i.6th 
power of the intensity of induction. Steinmetz # has 
expressed the hysteretic loss for transformers, obtained 
from experiment, by a formula similar to the following : 

W h =hfB™, (55) 

in which W h is the loss in watts per cubic centimeter 
due to hysteresis, h is the hysteretic constant which 

* Compare Steinmetz, Alternating Current Phenomena , Art. 98. 



MAGNETISM. 115 

varies with the quality of iron and steel from 20 x io -11 
to 10 x io~ 9 . Steinmetz found certain sheet iron trans- 
former plates to have h = 24 x io" 11 . Possibly a good 
average for transformer plates would be 21X10" 11 . f in 
the formula is the frequency of the alternating current 
and B is the intensity of induction in gausses. 

(k) Eddy or Foucault currents should properly be con- 
sidered here, since the effect is observed in iron cores. 
They are irregular currents developed in iron subjected 
to varying magnetic induction. Like hysteresis their 
energy is wasted in heating the iron. Steinmetz # has 
developed a formula for calculating the losses due to eddy 
currents in laminated iron, which may be stated thus : 

^ = Wx.o" 18 . (56) 

W e is the Foucault loss per cubic centimeter, / the thick- 
ness in mils of the laminations, or plates,/" the frequency, 
and B the maximum induction as in the hysteresis 
formula. 

The total loss in V cubic centimeters is therefore 

w e = r(p) 2 X io- 16 . ( S7 ) 

To express in watts per pound of iron we have 

W e = 6V(tfByx io" 15 . (58) 

Vis the volume in cubic inches. The eddy currrent loss 
is only 15% to 25% of the total core losses. 

* Reference cited, Art. 89. 



Il6 ELECTRICAL AND MAGNETIC CALCULATIONS. 



VIII. 

RELATION OF MAGNETIC QUANTITIES. 

37. The Law of Magnetic Force. — The definition already 
given for unit pole is one which will exert upon a similar 
one at a distance of 1 centimeter a force of 1 dyne. 

Example. — Develop a formula for the force exerted 
between two magnetic poles. 

Solution. — Let m and m' be the pole strengths and 
d the distance between them. First take the, distance 
1 cm. Then if one pole had unit strength and the other 
strength m, the force, from the definition would be 
F= m x 1 = m dynes. But the pole strength is rri in- 
stead of unity. Hence the force at 1 cm. will be 
m x m' = nim' dynes of attraction or repulsion, depend- 
ing upon whether they are unlike or like poles. Now 
since the magnetic field extends in all directions, the sur- 
face of influence will increase as the square of the distance 
away from the pole ; therefore the intensity of the force 
will decrease in the same ratio. If the distance is d, the 

force will be -^th of mm r \ or 

F=~^- (59) 

This law may be stated thus in words : Magnets attract or 
repel each other with a force proportional to the product of 



RELATION OF MAGNETIC QUANTITIES. 1 17 

their pole strengths and inversely proportional to the square 
of their distance apart. 

The forces due to a magnet pole at different distances 
may readily be determined by using a short piece of 
magnetized knitting needle suspended on a silk thread, 
obtaining the number of its swings in a given time at the 
different distances from the pole to be tested. It can 
readily be shown that the strength of field at any point is 
proportional to the square of the number of oscillations 
made by the needle at this point. Different pole strengths 
can be compared in the same way by taking the swings at 
the same distance from each. Of course the earth's field 
must be taken into account in these tests. The following 
examples will illustrate these points. 

Example. — Compare the earth's field intensity at a 
point A where the needle makes 24 vibrations a minute 
with its intensity at B where it makes 36 vibrations per 
minute. 

Solution. — Calling the horizontal intensity of the 
earth's field at the two places respectively H a and H h , 
from the principle stated we have 

H N* 

H=W < 6 °) 

N 

— - a is the ratio of the number of vibrations at A and B 

h 
respectively. Hence jf a J^ 2 4 

H h 36 s 9 

Example. — It is found that at the point A the needle 
makes 60 swings per minute when suspended 20 cm. from 



Il8 ELECTRICAL AND MAGNETIC CALCULATIONS. 

a certain long magnet which we shall call i ; it also 
makes 72 swings per minute when placed the same dis- 
tance from long magnet 2. Compare the intensities of 
the two magnet fields. 

Solution. — The forces in the two cases causing the 
vibrations are due to two things : the earth's field and 
the magnet field, since the needle was so arranged with 
respect to the magnet and the earth as to be in the 
magnetic meridian passing longitudinally through the 
magnet. Let H represent the earth's field intensity, and 
M x and M 2 respectively the magnet fields together with 
the earth's field. Also let JVbe the number of swings in 
the earth's field alone, and JV t and N 2 be the numbers 
due to the total field in the two cases. Hence 

F[ _ M x - H __ N 2 - N 2 
F 2 ~ M 2 - H~~ N 2 - N 2 ' 

Here F ± and F 2 represent the forces due to the magnets 
alone. For the particular values given, therefore, 

F x _ 60 — 24 21 
-^2 72 2 — 24* 3 2 

The intensity of field due to the first magnet pole at 
20 cm. is § \ of that due to the second magnet pole at 
the same distance. 

Example. — At what distance apart are two magnet 
poles when a mutual force of 500 dynes is measured 
between them, the poles having 100 and 500 units of 
strength respectively ? 



RELATION OF MAGNETIC QUANTITIES. Iig 



Solution. — 



„ mm! 100 X 500 

^=-^=—^ = 500. 



Hence 



4 /ioo X coo 
a = y — - = 10 centimeters. 



500 

Example. — Two poles of equal strength are 20 centi- 
meters apart and the force of mutual attraction is 2.25 
dynes. Find the strength of the poles. 

Solution. — From the above formula 

mm! = Fd 2 = 2.25 x 20 = 900. 

But m = m! ; hence m 2 = 900, and m = 30. Also m' — 30. 

Example. — Show that the following experimental data 
prove that the force due to a magnet varies inversely as 
the square of the distance from the pole. The small mag- 
netic needle used made 15 swings in a minute in the 
earth's field alone, and when suspended in the meridian 
of the earth and a bar magnet at a distance of 20 centi- 
meters from the latter it made 2 5 swings in a minute ; 
when placed 40 centimeters from the magnet pole it made 
18 swings in a minute. 

Solution. — 

F[ N 2 -N 2 ^ 2 -i? 4 
F 2 N 2 -N 2 78 2 -i^ 2 "i 
The force at 20 cm. is four times the force at 40 cm. 
while the distance is \. Therefore 



— =2 = 4> 



■^2 "1 20' 
or the forces are inversely as the squares of the distances. 



120 ELECTRICAL AND MAGNETIC CALCULATIONS. 

Example. — The vibrating needle swinging in the 
earth's field makes 10 vibrations a minute; the #-pole 
of a magnet which is placed in the magnetic meridian is 
brought near the ^-pole of the needle when the latter 
makes 20 vibrations a minute. How much stronger 
field does the magnet produce at the given distance than 
the earth ? 

Solution. — 

F 1 N 2 -N 2 W-7? 



F~ N 2 io 2 = 3 



The magnet's field is 3 times as intense as the earth's 
field. 

38. Magnetic Flux, Intensity of Magnetism and Strength 
of Field. — Conceive a spherical surface, radius one cen- 
timeter, drawn about a pole whose strength is 150 
units ; how many maxwells of flux will pass through the 
surface ? 

Solution. — The surface is 47r sq. cm., and since the 
pole is 150, the density on the surface will be 150 lines 
per sq. cm. Hence 

<£ = 47T X 150 = 1885 maxwells. 

Example. — What must be the intensity of magnetism 
when a magnet of 20 sq. cm. cross section is excited to 
produce 200,000 maxwells? Find /jl from the table of 
permeabilities. 

Solution. — From the formula given 

_ <f> 200,000 

B — — = = 10,000 gausses. 

A 20 



RELATION OF MAGNETIC QUANTITIES. 121 

Hence /x for annealed iron = 2000. 

Example. — What is the intensity of a magnetic field 
when a bar of soft iron which is placed in it has an 
induced intensity of 16000 gausses at a permeability 
of 300 ? 

Solution. — Since B = y,H, 

B 16.000 
H = — = = 53.3 gausses. 

fx 300 

Example. — What is the permeability of a piece of soft 
iron made into a ring so that a field of 100 gausses pro- 
duces an induction, or intensity of magnetism, of 18000 
gausses ? 

Solution. — From B = y,H, 

B 18,000 

H= — = = 180. 

H 100 

The relation of B and H may be experimentally deter- 
mined by what is known as the permeameter method* A 
rectangular piece of iron has a slot cut through it in which 
the magnetizing coil is placed. The coil has a cylindrical 
hole through it longitudinallly to receive the rod to be 
tested. The iron is drilled above the coil to permit the 
rod's being placed in the coil. The lower end of the rod 
is surfaced to make good contact with the iron below the 
coil. When the current is turned on the attractive force 
causes the rod to stick so that a force is necessary to pull 
it out, which is measured by a spring balance. 

Suppose the coil to be 12 cm. long and have 300 turns, 

* See Thompson, Lecttires on the Electromagnet, p. 70. 



122 ELECTRICAL AND MAGNETIC CALCULATIONS. 

thus making 25 turns to the centimeter. The intensity of 
field, or magnetizing force, is then, per ampere of current, 

H= 1.25 x nl= 1.25 x 25 x 1 = 31.25 gausses. 

If 10 amperes be used, 

H= 31.25 x 10 =312.5 gausses. 

Now suppose the area of contact, or end of the rod, be 
A sq. in., and the pull to detach it be W pounds. Then 
the formula for B, whose derivation is given below, is 



B= 1317 



\j^ + H. (62) 



Example. — If A is in square centimeters and W is 
in grams, what will the formula become ? 

Solution. — Since 1 cm. = f in. and 1 in = f cm., 
1 sq. in. is then - 2 ¥ 5 - sq. cm. ; also, since 1 lb. = 453.6 
grams, the substitution of the proper constants gives 

' = w v/^f- 6 + *■ 

Taking the square root of the constants under the radical 
sign and placing the result outside, this reduces to 



B 



= i S 6sJ^+JT. (63) 



This is to be used when A is sq. cm., and W is grams to 
pull the rod out. 

Example. — To derive this last formula from Maxwell's 
formula for the lifting power of magnets ; namely, 

W '= — — ? as explained later. 

981 x 8tt 



RELATION OF MAGNETIC QUANTITIES. \2$ 

Solution. — In this W is the number of grams of 
weight lifted, A is the number of square centimeters of 
contact of the magnet pole with the armature or keeper. 
Transposing this, 

JVX 981 x'8tt 

™ = —. 

A 

Whence , j~w fw 

b = V981 X 8^xy— =i S 6y-^- 

When the induction is caused by a field of H gausses, 
this becomes „ , fw 

5 \~A +H - 
Example. — Find the permeability of a sample of char- 
coal iron when a test by the permeameter described gave 
the following data : coil carries 20 amperes, area of rod 
tested is 1 square centimeter, and it requires a force of 
5794.25 grams to pull the rod out. 

Solution. — First obtain B. 

B = T56 1 / 5794- 2 5 -f- 37. 2 5 x 20 = 12,500 gausses. 

Therefore B 1 2 , c o o 

fl =-— = — = 20. 

H 625 

39. Ohm' 8 Law and the Magnetic Circuit. — As the 

magnetic circuit is analogous to the electric circuit, so 
the law governing the magnetic flux is analogous to Ohm's 
law governing electric currents. Expressing the amount 
of flux by <j>, the magnetomotive force by M, the reluctance 
by (ft, we have J/ 

* = ^- (64) 



124 ELECTRICAL AND MAGNETIC CALCULATIONS. 

Example. — How many oersteds of reluctance will there 
be in a wooden ring whose mean circumference is ioo 
centimeters and whose cross section is 20 square centi- 
meters ? 

Solution. — Since (R = k- > and in wood k = 1, 

a 

^ IO ° 

61 = 1 X = «; oersteds. 

20 

Example. — Suppose the wooden ring wound uniformly 
with 1000 turns of insulated wire and 10 amperes of cur- 
rent sent through it ; what will be the M.M.F. in gilberts 
and also the magnetizing force, or field intensity ? 

Solution. — ;//= 1000 Xio= 10,000 ampere-turns. 

Also M= 1.256 X nl = 1.256 x io 4 = 12,560 gilberts. 

H — — = 125.6 gilberts per cm. or gausses. 

Example.— Show how M = = i.2z6nl. 

10 ° 

Solution. — Assume unit pole within a long helix 
having n turns to the centimeter and carrying / amperes. 
Now let the pole be moved along the axis of the coil 1 
centimeter, when each of the \tt lines of force from the 
unit pole will be intercepted by n turns of wire, or the 
lines cut will be 47m ; and since there are / amperes 
the work done in moving the pole is 47ml, in which / is 
in absolute amperes. But work is Fl, or in this case ZT/, 
where Zf is the magnetizing force. Hence 

HI = 4.7m/; 



RELATION OF MAGNETIC QUANTITIES. 12$ 

but /= i, therefore H '= ^-khI, in this particular instance. 
If, however, the pole had been moved the whole length 
/ of the solenoid, the whole work done is 

H/ = 41ml =M, (65) 

since M = HI, or the gilberts-per-centimeter times the 
number of centimeters in the coil. Now if /is in inter- 
national amperes, 

\TTtlI 

M= = 1.216 nl. 

10 ° 

Example. — How many maxwells of flux will pass 
around the ring above described under the conditions 
given ? What is the intensity of magnetization ? 

Solution. — d> = — — = — — — = 21:12 maxwells. 

<R 5 D 

<t> 2 C12 
Also B = — = — — = 125.6 gausses. 

It is seen that B here is the same as H above ; this is 
correct as there is no iron in the core and /x = 1, so B 
= !*,&-= H. 

When the magnetic circuit contains iron the reluctance 
is not so readily found, since the reluctivity k varies 
with the intensity B. It may, however, be determined 
satisfactorily if the quality of iron is definitely known. 
The following constants from Houston and Kennelly * are 
here given for some kinds of iron most commonly met 
in practice. 

* Electro-dynamic Machinery, Art. 68, p. 65. 



126 ELECTRICAL AND MAGNETIC CALCULATIONS. 

For ordinary dynamo cast iron, 

k = 0.0026 + 0.000093 If. (66) 

For dynamo wrought iron, 

k = 0.0004 + 0.000057 If. (67) 

For soft iron, according to Stoletow, 

k = 0.0002 + 0.000056 If. (68) 

For cast iron, 

k = 0.0010 + 0.000129 If. (69) 

For Norway iron, 

k = 0.0001 + 0.000059 H. (70) 

For steel, k == 0.00045 + 0.000051 If. (71) 

These equations will give the reluctivity k of the kinds 
of iron mentioned for different intensities of field H. 
They apply, however, to closed magnetic circuits and are 
inaccurate for circuits having air gaps. 

Example. — Suppose the ring under consideration 
were of dynamo wrought iron, what would be its reluc- 
tance, the magnetic flux, the intensity of magnetization 
and the permeability ? 

Solution. — The field as previously found is If = 
125.6 gausses; hence 

k = 0.0004 + 0.000057 X 125.6 

= 0.00758 oersted per cc. 

100 

Hence (R = 0.007^8 X = 0.0370 oersted. 

■ 20 

M 12,560 „ 

Also <t> = -r- = — - — = 331,400 maxwells. 

(R 0.0379 



RELATIONS OF MAGNETIC QUANTITIES. \2J 



Therefore B = -^ = 33I ' 4QO = 16,570 gausses. 

B 16,570 

,, = 1 = — i-^ = i 32 , as before. 

For accuracy, so far as this is possible in the calcula- 
tions of magnetic circuits where they are partly iron and 
partly air, as in horse-shoe magnets and in dynamos of 
all kinds, we shall have to find k for the metal portion in 
terms of the induction B in the metal instead of in 
terms of H. The reason is that though we may have the 
same intensity of field in the case of an open circuit as in 
the case of the closed ring, yet the total reluctance being 
greater, the induction in the iron is less ; hence we should 
obtain a value too large for k by using the above formula 
for open magnetic circuits. Hence a formula which will 
give k directly in terms of B itself will be more nearly 
correct. 

In the formula for k, call the first constant a, the sec- 
ond b ; the general equation is then 

k = a + bH. (72) 

We also have, from analogy to the electric circuit, 

*=? = ** ' 

Whence H=Bk; 

substituting above, k = a + bBk. 

Hence £ = _^__. ( 73 ) 



128 ELECTRICAL AND MAGNETIC CALCULATIONS. 

Example. — Suppose the iron ring in the last example 
have a gap i centimeter wide cut through it in one place ; 
what effect will this have on the total reluctance and the 
total lines of force in the ring ? 

Solution. — First the reluctance of the air gap is 

(R a = - = — =0.0 k oersted. 
a 20 ° 

The reluctance of the iron portion is also 

a 20 

But k is to find. To use the formula just worked out for 
reluctivity in terms of B, it is observed that we must 
know B first. Now the air gap adds 0.05 oersted to the 
reluctivity of the iron which before was 0.0379, hut which 
will be less now because the induction is less. So the 
total reluctance is seen to be over twice the reluctance 
before the gap was made. Approximately it is twice. So 
the flux will be approximately \ of the former, giving an 
intensity of induction B about 8000 gausses. There- 
fore 

a 0.0004 

i — bB 1 — 0.000057 x 8000 
= 0.00074 oersted per cc. 

Substituting in the equation for iron reluctance, 

(R. = 0.00074 x f -jy = 0.0036 oersted. 

The total reluctance is 

(R a + (R. = 0.05 + 0.0036 = 0.0536 oersted. 



DELATION OF MAGNETIC QUANTITIES. 1 29 

Hence the total flux is 

12,560 
d> = — ^—z = 234,000 maxwells. 
0.0536 

_. 234,000 
B = — = 11,700 gausses. 

This is somewhat larger than that assumed in obtaining 
k, but the slight difference this will make in the iron 
reluctivity will not materially affect the total reluctance, 
since the air gap is so large a proportion. 

For example, take 11,700 = B and apply in the 
formula for k, whence 

_ 0.0004 

k = = 0.00090, 

1—0.000057x11,700 

which will not materially alter the iron reluctance, and 
much less the total reluctance. 

The problem is most often one to find the necessary 
M.M.F. to supply a given flux in the air gap or through 
the magnet core. In this event the reluctivity at the 
corresponding flux density may readily be found from 
(73) when the quality of the iron is known. 

Example. — Assuming that there are 1000 turns on 
the ring, how many amperes wall be necessary to provide 
the M.M.F. as required in the previous example ? 

Solution. — For the air gap, if it carries 234,000 
maxwells, 

M a = <£ a (R a = 234,000 x 0.05 = 11,700 gilberts. 



13O ELECTRICAL AND MAGNETIC CALCULATIONS. 

For the iron portion, assuming the reluctance as 
estimated, 

M { = cf> { (R,. = 234,000 x 0.0036 = 842 gilberts. 

Therefore the total magnetomotive force is 

M= 11,700 + 842 == 12,542 gilberts. 

But M = 1.256 nl. 

Therefore 

M 12,1:42 . , 

nl = = — ^— r-= 10,000 amp.-turns, approximately. 

1.256 1.256 

Hence _ 10,000 

I — — = 10 amperes. 

1000 

Example. — It is desired to find the M.M.F. in gilberts 
and the number of ampere-turns to excite the ring so that 
12,000 gausses may be the air-gap density, assuming no 
leakage. What will be the permeability of the iron under 
these conditions ? 

Solution. — We have first to find the reluctance. 
The reluctivity of the iron is 

a 0.0004 

1 — bB 1 — 0.000057 x 12,000 
= 0.00126 oersted per cc. 

The total reluctance is therefore 

(R = — + k- = h 0.00126 x — = 0.056237 oersted. 

a a a { 20 20 

Hence 

M = </>(R = 12,000 x 20 x 0.05624 = 13,497 gilberts. 



RELATION OF MAGNETIC QUANTITIES. 13I 

M 13,497 

Also nl = = — — v- = 10,746 amp.-turns. 

1.256 1.256 

1 1 

/* = T = ^^ T ^ = 794. 
k O.OOI20 

We may also obtain H, the intensity of field, from 

Example. — Assume a coefficient of leakage of 1.2. 
Determine the values of the quantities as in the last 
example. 

Solution. — The flux desired in the air gap, as before, 
is 12,000 x 20 = 240,000 maxwells. The necessary flux 
in the ring is 

240,000 x 1.2 = 288,000 maxwells. 

_, 288,000 

B = = 14,400 gausses. 

20 

0.0004 . 

k = = 0.002232 oersted per cc. 

1 — 0.000057 X 14,400 

Therefore 

(R. = 0.0c + 0.0022^2 x — = 0.0610c: oersted. 
° 20 ° 

M= <£(R=(i4,4oo x 20) x 0.06105 = I 7?S^° gilberts. 

ai t 17,^80 

Also nl = „ = 14,000 ampere-turns. 

1.256 

This is an increase of 3254 ampere-turns over that in the 
last example, caused by leakage. If the ring is wound 
with 1000 turns the wire must be taken large enough 
to carry 14 amperes of current. 



132 ELECTRICAL AND MAGNETIC CALCULATIONS. 

It may seem that we should take only 12,000 gausses 
for the intensity in the air gap, while 14,400 is the inten- 
sity in the iron. But it takes M.M.F. to carry the leak- 
age lines through the air as well as the useful ones, and 
to consider that only 240,000 lines passed through the air 
will obtain an M.M.F. too small. Although we do not 
know the exact path of the leakage lines, our results will 
be the more accurate if we consider that the whole flux 
set up in the core passes across the air space and obtain 
the M.M.F. for the latter accordingly. 

40. The Lifting Power of Magnets. — As already given* 
the formula of Maxwell for the tractive force of magnets 
for their armatures when there is no appreciable air gap 
at the contact is TTr IP A , x 

W = T^' (74 > 

in which W is the weight lifted in dynes, and A is the 
area of polar contact in square centimeters. To reduce 
this to grams lifted divide by 981 and the formula 
becomes TTr B 2 A , N 

To express the tractive force in pounds divide again by 
453.6 ; whence, reducing the denominator, 

W= p* ,. ( 7 6) 

11. 183 X io 6 v/ J 

It is to be observed that horseshoe magnets, such as are 
used in factories, machine shops, etc., for heavy lifting, 
have a double area of contact. Hence A in the formulae 
is the sum of both polar surfaces. 

* See Equation (51) p. kn ; also see p. 122. 



RELATION OF MAGNETIC QUANTITIES. 1 33 

Example. — An oblong horseshoe magnet has a cross 
section of 20 square centimeters, and the keeper, or 
armature, has the same shape and the same cross section, 
and its ends fit very closely to those of the magnet 
proper. The magnet is excited to an intensity across the 
plane of contact of B = 15,000 gausses. How many 
dynes, grams and pounds can be supported ? 

Solution. — 

TTr B*A (i5,ooo) 2 X 20 X 2 

W = —— = ° J = 358,097,000 dynes. 

07T O X 3.I4IO 

Also 

B*A (i5,ooo) 2 X4o , 

JV=- — = — — ^~- = 365,000 grams. 

8tt x 9 81 25.13x9 s ! 



W= Q ^ 6 = 805 lbs. 



And m B*A 

11. 183 x io ( 

The values, of course, include the weight of the armature. 



Example. — Let the total length of the iron circuit, 
magnet and keeper, be 80 cms. What M.M.F. will be 
necessary to magnetize to 15,000 gausses, and if the 
exciting current is to be 5 amperes, how many turns of 
wire must be put on ? The magnet is made of dynamo 
wrought iron. 

Solution. — </> = BA = 15,000 X 20 = 300,000 max- 
wells. The reluctivity is 

. a 0.0004 



\—bB 1 — 0.000057x15,000 
= 0.0028 oersted per cc. 
The reluctance is, using a instead of A, 

(R = k— = 0.0028 — = 0.0112 oersted. 
a 20 



134 ELECTRICAL AND MAGNETIC CALCULATIONS. 

Hence Jlf= <£(R = 300,000 x 0.0112 = 3360 gilberts. 

Also _ 3360 

nl = =2675 amp.-turns. 

\j 

2671: . . 

n = — — = 535 turns of wire. 

Example. — It is not economical to excite magnets 
much beyond 16,000 gausses. Show to how many pounds 
per square inch this is equivalent. 

Solution. — For pounds per square centimeter we 
have from (76) 

us B (i6,ooo) 2 

TV= r = — ^— - y — s = 22.9 pounds. 

n. 183 x io 6 n.183 x io 6 y 

One sq. in. = 6.45 sq. cm. Hence per sq. in., 

B 2 X 6.45 

W= z -^-s = 147.6 lbs. 

11. 183 x io 6 ' 

Thompson gives the economical limit of 150 lbs. per 
square inch. This will be equivalent to an induction of 

D 4 /150 X 11. 183 X IO 6 
B = y — = 16,120 gausses. 

For traction purposes we may, in general, figure on 150 
lbs. to the square inch, or about 24 lbs. per square 
centimeter. For cast iron B should not be over 6,000 or 
7,000 gausses, and W about 28 lbs. per square inch. 
These values given, we may find at once the required 
cross section for any given weight to be supported. 

Example. — Let it be required to design a horseshoe 
magnet to carry 2 tons. 



RELATION OF MAGNETIC QUANTITIES. 1 35 

Solution. — 2 tons = 2 x 2240 = 4480 lbs. 

A = 4480 -5-150 = 30 sq. in. ; or 15 sq. in., polar area. 

Since the magnetic circuit should be as short as possi- 
ble consistent with sufficient winding space, we shall 
take the length of magnet to be 20 inches, a straight 
armature 14 inches long, and assume the material to be 
Norway iron. Whence 

a 0.0001 

1— bB 1—0.000059X16,000 

= 0.0018 oersted per cc. 

^ t / uX 2.C4 

(31 = k- = 0.0018 X — — y- 2 - 1 = 0.0016 oersted. 
a 15x6.45 

Hence 

M •=■ <£(R = 0.0016 x 16,000 x 96.7 = 2475 gilberts. 

M 247c 

nl = - = — -^ = 1080 amp.-turns. 

1.256 1.256 

Example. — Suppose storage batteries capable of giv- 
ing 5 amperes are used to excite the above magnet. 
Determine the length, size and resistance of the wire to 
be used. Also approximate the space to be allotted 
to the winding. 

Solution. — Thompson gives ^ inch as the maximum 
depth of winding for ordinary magnets, and if this be 
reached only about 1 inch of winding space need be used 
for each 20 inches of the iron circuit. In this case there 
are required 

1980 -r- 5 = 400 turns, approximately. 



136 ELECTRICAL AND MAGNETIC CALCULATIONS. 

The wire must be large enough for 5 amperes without 

overheating. From the table of safe carrying capacities 

we find No. 16 wire is sufficient. This has 50.8 mils 

diameter bare, or about 67 mils double cotton covered. 

Take winding space 2 inches = 2000 mils. In one layer 

there will be 

2000 -=- 67 = 30 turns. 

Hence there must be in depth 

400 -5- 30 = 13 layers. 

This gives a depth of 

13 x 67 = 871 mils = 0.87 inch, 

which is above Thompson's limit. Hence to reduce depth 
we must allow more space, say 3 inches' = 3000 mils. 
In one layer there will be 

3000 -f- 67 = 45 turns. 
And 400 -5- 45 = 9 layers deep. 

The depth of winding now is 

9 X 67 = 603 mils = 0.6 inch. 
The average length of one turn is 



(Vi5-^o. 7854 + 0.6) x 3.1 416 = 15.7 inches, 
Total length will be 

15.7X400 r 

*' =523 feet. 



12 

The resistance will be 



R = kL= IO>79 x _g|_ = 2>l8 ohms . 

It has been found that however closely the armature fits 
against the poles the reluctance of the contact is still 



RELATION OF MAGNETIC QUANTITIES. 1 37 

equivalent to 0.003 to 0.004 centimeter of air. Take 
0.0035 centimeter as a fair average. 

Example. — Determine the M.M.F. for the magnet con- 
sidered above, taking account of the contact reluctance. 

Solution. — The reluctance of. the equivalent air 

space is 

/ 0.003c 
(R = k- = — -^ = 0.000036 oersted. 
a 96.7 

Therefore the total reluctance is 

(R = 0.0016 + 0.000036 x 2 = 0.0017 oersted. 
M= <£(R = 16,000 x 96.7 X 0.0017 = 2630 gilberts. 
nl = 2630 -r- 1.256 = 2100. 

This makes an addition of 

2100 — 1980 = 120 ampere-turns, 

due to the contact reluctance, or if £■ = 24 turns of wire 
additional. We might neglect this slight increase in the 
reluctance in the calculation by slightly increasing the 
exciting current. In this example the current would need 
to be increased ^ ampere. 

41. Temperature and Magnetism. — If a wire nail be 
magnetized, then heated to redness, tests will show that it 
has thus lost all trace of its former magnetism. Experi- 
ment is also able to determine the rate at which the 
intensity of magnetism decreases with the rise of tempera- 
ture. Let a be the decrease in the magnetic moment of 
a magnet (which is proportional to the magnetic strength) 
per degree of temperature per unit of magnetic moment ; 
/ the rise of temperature, M Q the magnetic moment at o° C 



138 ELECTRICAL AND MAGNETIC CALCULATIONS. 

The decrease is then M Q x at. Hence the magnetic 
moment M t at any increase of temperature / will be 

M t = M Q -M x at= M (i - at). (77) 

This formula is similar to that for the resistance of a 
wire at any increase* of temperature above that at which 
the resistance is given, except that the temperature co- 
efficient of resistance is usually positive, while the tempera- 
ture coefficient of magnetism a as given in the formula 
(77) is negative. This coefficient is small, so that a con- 
siderable change of temperature is necessary to ap- 
preciably affect the strength of the magnet. 

Example. — The magnetometer needle stands in the 
magnetic meridian ; a small 4-inch magnet is placed in an 
oil bath in an east-west position at a short distance west 
from the center of the magnetometer needle. The tem- 
perature of the oil bath is 20 C. and the deflection of the 
magnetometer 150 millimeters after the magnet has ac- 
quired the temperature of the oil. The oil is now heated 
slowly to 50 C. and the deflection becomes 148. Find 
the temperature coefficient of the magnet. 

Solution. — The magnetic moment M—ml\ i.e. the 
pole strength times the length, and hence the pole 
strengths m are proportional to the deflections. Hence 
from (77) 

148 = 150 [1 - a (50 - 20)] = 150 - 4500 a. 

Therefore 

148 — ICO 

a = — = 0.00044. 

— 4500 



RELATION OF MAGNETIC QUANTITIES. 1 39 

42. Hysteresis and Eddy Currents. — Example. — 
Compare the hysteresis losses in two transformers having 
the same induction but in one of which the frequency is 
60 and in the other it is 120. 

Solution. — 

Wj! tifB r ™ f_ _6o 1 
W»" ~ h"f"B"™ ~ /" " 120 " 2 ' 

since h! = h" and B r = B" . That is, since the loss varies 
according to the frequency, the first will have one-half the 
loss of the second. 

Example. — Find the watts lost due to hysteresis in a 
transformer of 4000 cubic centimeters in which B — 3000 
gausses, f = 100, and the hysteretic constant h = 20 x 
io~ n watts per cubic centimeter per cycle. 

Solution. — The watts lost per cc. are given in the 
formula 

JV h = hf&*, 
and for V cu. cm., 

VW h = VhfB™= 4000 X 20 X 10- 11 X 100 X ^W 6 

o — «i 1-6 

= 8 x 10 5 x 3°°° • 

We find now from a table of logarithms the logarithm of 
3000, multiply it by 1.6 and find the number correspond- 
ing; this gives the 1.6th power of 3000. Proceeding, 

3000 = 100 x 30, 

and log3ooo = log 100 + log3o = 2 +1. 47712 = 3.47712* 

3.47712 x 1.6 = 5.56339 

which is found to be the log of 365924; the latter i? 



14O ELECTRICAL AND MAGNETIC CALCULATIONS. 



1.6 



therefore the i.6th power of 3000; or 3000 = 365924. 
Substituting above, we have 

V1V h = 8 x io~ 6 X 3.65924 x io 5 = 29.27 watts. 

Example. — Find the watts lost in the transformer con- 
sidered due to eddy currents if the plates, or laminae, are 
12 mils thick. 

Solution. — The formula for eddy currents is 

W e = V(t/Bf x io" 16 . 
Therefore 

W e = 4000(12 x ioox3ooo) 2 x io -16 = 5.18 watts. 

Example. — If a certain test of this transformer shows 
6 watts lost due to eddy currents, what is the induction ? 

Solution. — From the above formula, 



B 



V Fx(i2 XiooVXio- 16 V , 



rx(i2Xioo) 2 Xio" 16 V 4oooX(i2oo) 2 X io -16 
= 3227 gausses. 

43. Original Problems. — 1. If a magnetic needle make 
24 oscillations per minute at a point where the horizontal 
intensity of the earth's field is 0.2 dyne, what is the hori- 
zontal intensity at a place where the same needle makes 
22 oscillations per minute? H= 0.168 dyne. 

2. A given magnetic needle makes 24 vibrations in the 
earth's field alone. When magnet A is placed in the 
meridian at 20 centimeters from the needle, the latter 
makes 31 vibrations per minute. When magnet B is 
placed in the same position and at the same distance the 
needle makes 28 vibrations per minute. Compare the 
forces due to A and B. A = 1.85 B, 



RELATION OF MAGNETIC QUANTITIES. 141 

3. Compare the pole strength of A, 30 swings per 
minute, with that of B, 36 swings per minute, when the 
needle makes 24 swings per minute in the earth's field/ 

B 20 

4. What is the magnetic density when a spherical sur- 
face whose radius is 5 centimeters has at its center a 
magnetic pole whose strength m =■ 200? How many- 
maxwells of flux pass through this surface ? 

B = 8 gausses. 

<f> = 2512 maxwells. 

5. A ring of annealed iron is required to have 600,000 
maxwells of flux at a density of 15,000 gausses. What 
must be its cross section, and what is its permeability ? / 

A = 40 sq. cm. 

fx= 500. 

6. What is the field intensity in the last problem, and 
how many amperes will be necessary to produce the field 
if the coil has 50 turns per centimeter ? 

H= 30 gausses. 
/= 0.48 ampere. 

7. The cross section of the rod used in a permeameter 
for testing permeabilities is 1 sq. cm., and requires a force 
of 10,000 grams to pull it off when the coil having 25 
turns per centimeter carries 10 amperes. What is the 
permeability of the test piece ? /x = 50. 

8. If the rod is required to have a permeability of 30, 
how many grams of force will pull it off when the current 
is 15 amperes? W= 8000 grams, 



142 ELECTRICAL AND MAGNETIC CALCULATIONS. 

9. Determine the reluctance of a brass ring whose cir- 
cumference is 125 centimeters, and cross section 2 sq. 
cm., if an air gap of 2 cms. length be cut across it. 

(ft = 62.5 oersteds. 

10. What M.M.F. will be required to produce a density 
of 200 gausses in the ring, and how many ampere-turns 
must be wound on the ring for this purpose ? 

M= 25,000 gilberts. 
n/= 20,000 amp.-turns. 

1 1 . What will be the field intensity if the ring is wound 
for 25 amperes of current? U= 200 gausses. 

12. Suppose the above ring were of wrought iron and 
have no air gap. Find its reluctivity, reluctance, flux 
density and permeability. k = 0.0118 oersted per cc. 

(ft = 0.7375 oersted. 
B = 16,800 gausses. 
fx= 84. 
*i3. How many turns of wire for 2 amperes must be 
wound uniformly on this ring to produce a flux of 20,000 
maxwells ? What will be the field intensity under these 
conditions ? n = 465 turns. 

//= 9.3 gausses. 

14. Suppose 1 cm. of the ring be cut out, thus forming 
an air gap. How many turns of wire for 5 amperes ex- 
citing current must be wound on to give an intensity of 
10,000 gausses in the iron and 9090 gausses in the air 
gap, thus making the leakage coefficient 1.1 ? 

n = 1784 turns. 

15. Find the permeability of the iron in the last 
problem. ft = 1075. 

* J5ee note on p. 147. 



RELATION OF MAGNETIC QUANTITIES. 1 43 

*i6. A magnet of square horseshoe shape has the fol- 
lowing dimensions : total length of iron 60 cms. ; cross 
section 25 sq. cms. ; length of air gap 0.075 cms - '-> area 
of air gap 25 sq. cms. The armature is drum, or cylin- 
drical, in shape as in motors and dynamos, and is 7 cms. 
in thickness from air gap to air gap. The magnet is 
wound with wire for 2.5 amperes to furnish a flux of 
350,000 lines available through the armature and air 
gaps, coefficient of leakage estimated at 1.2. How many 
turns of wire will it take and what winding space will be 
necessary, allowing 1000 circular mils per ampere and 
25% for insulation of wires, and permitting a depth of 
wire of 1.5 inches ? Also how many feet of wire will be 
required? For magnet M.M.F. = 5000 gilberts. 

For armature M.M.F. = 35 gilberts. 

For air gap M.M.F. = 2520 gilberts. 

Total M.M.F. = 7555 gilberts. 

n = 2418 turns of wire. 

Wire is 2500 cir. mils = No. 16 =50.8 mils bare = 63 
mils covered. Wire space = 6.3 inches. 

Length of wire = 2829 ft. 

17. How many dynes, grams and pounds may be sup- 
ported on the hook of a lifting magnet having the follow- 
ing constants : induction 16,000 gausses across polar con- 
tact ; weight of armature 20 lbs. ; area of pole faces 25 

sq. cms. each ? „, 

W = 5.12 x io 8 dynes. 

= 521.7 kilos. 
= 1150 pounds. 
From these subtract the weight of the armature. 

* Work by curves, p. 229, or tables, p. 298. 



144 ELECTRICAL AND MAGNETIC CALCULATIONS. 

1 8. Assuming that 16,000 gausses is a fair limit to 
economical induction, what must be the area of a magnet 
to support 1 ton (2000 lbs.) including the weight of the 
keeper ? A = 43.68 sq. cms. 

= 6.77 sq. ins. 

19. If a test shows that a lifting magnet is carrying 
1600 lbs., while its area of contact is 10 sq. ins., what 
must be the induction across the area of contact ? 

B = 11,800 gausses. 

*2o. If the total length of the iron circuit in problem 
1 7 is 1 2 inches and the material is dynamo wrought iron, 
how many ampere-turns will be necessary to give the 
proper excitation? By curve, nl ' = 1391 amp.-turns. 

21. If a 6-ampere exciting circuit is available, how 
many turns and what size of wire will be necessary for 
the magnet in problem 18, if we may assume the total 
length of iron to be 20 inches? 

By curve, n = 380 turns of No. 12. 

22. How much space must be provided for the wire 
above, if we may allow a depth of 1 inch? Also what 
will be the length of wire ? Wire space = 3.8 inches. 

Length = 386 feet 

23. Required to design a horseshoe lifting magnet of 
circular cross section to carry 3 long tons. Specify all 
values for the construction of the magnet of Norway iron, 
taking a current of 5 amperes. 

• The answers in this and the next problem cover allowance for air gap. 



RELATION OF MAGNETIC QUANTITIES. 145 

Area = 22.4 sq. in., cross section. 
Length = 32 inches = 80 cms., estimated. 
Diam. = 5.34 inches = 13.5 cms. 
Depth of wire = 0.5 inch. 
Size of wire = No. 14 = 80 mils d.c.c. 
Length of wire = 735 feet. 
Winding space = 6^ inches total. 
Ampere-turns = 2400; n = 480 turns. 




24. Find the length and the resistance of the wire in 
the last problem if allowance be made for the reluctance 
.of the contact area. Also determine the depth of wire if 
the same space be allowed as before. 

nl = 2490 ; n = 498. 

L = 763 f t. ; R = 2.2 ohms. 

Wire is No. 14 B. & S. 

25. Give the design for a lifting magnet to be used 
in a factory where weights up to 1200 pounds are to be 
handled, constructing it of dynamo iron, allowing for 
joints, and having 3 amperes for excitation. Use tables. 



146 ELECTRICAL AND MAGNETIC CALCULATIONS. 

L = 40 cm., total; L = 28 cm., horseshoe. 
A — 25.8 sq. cm. == 4 sq. in. 
#/= 2012 ; n = 671 ; wire = No. 16 d.c.c. 
Depth = 8 layers = 0.5 in. ; space = 5.4 in., 

or 2.7 in. each side. 
L of wire = 526 ft; R = 2.7 ohms; wt. 

iron = 8 kilos. 

26. A steel magnet immersed in an oil bath at 15 C. 
at a certain distance from the magnetometer causes a 
deflection of 200 mm. Assuming its temperature coeffi- 
cient to be 0.00045, to what temperature must it be heated 
theoretically to lose £ its magnetism ? 

'= 570.5° c. 

27. Find the temperature coefficient of a magnet if 
when placed at a certain distance from the magnetometer, 
its axis on an east and west line, it gives a deflection of 
200 mm. at 20 C, but when heated to ioo° C. it pro- 
duces a deflection of 192 mm. 

a = 0.0005. 

28. How much greater is the loss in a transformer for 
a frequency of 120 with an induction of 4000 gausses 
made of iron such that the hysteretic constant is 21 x 
io~ n , than in one of the same capacity and with the 
same induction, for a frequency of 100, and having a 
hysteretic constant of 20 x io" -11 ? 

Hysteresis loss first = § % of second. 

29. Determine the total hysteresis loss of a transformer 
in watts, its volume being 6000 cc, induction 3200 
gausses, frequency 120, hysteretic constant 20 x io~ u 
watts per cc. per cycle. 



RELATION OF MAGNETIC QUANTITIES. 1 47 

VW h = 58.42 watts. 
Suggestion. — log 3200 = log 100 + log 32 

= 2 +1. 50515. 
log* = 1.6 x 3.50515 = 5.60824. 
Use in AfB 1 *, B 16 = x = 32oo 16 = 40573°- 

30. Calculate the eddy current loss in the above trans- 
former, the laminae being 10 mils thick. 

W e = 8.85 watts. 

31. With a certain exciting current this transformer 
experienced an eddy current loss Qf 8 watts. What was 
the intensity of induction ? B — 3043 gausses. 

32. Find the total core loss in a transformer of 5000 
cubic centimeters, frequency 100, induction 3000 gausses, 
hysteretic constant of the iron 20 x io~ n , thickness of 
plates 13 mils. 

W h — 36.59 watts. 
W e = 7.6 watts. 
Total = 44.2 watts. 

Note. Instead of the more tedious reluctance method of calcu- 
lating the required ampere-turns for a magnetic circuit, as described 
in this Chapter, and originally used in the solution of the problems 
in this section, it will be more convenient to use the curves given on 
page 229. Obtain B from the given conditions, find H correspond- 

Hl 

ing from the proper curve, whence ni — = .8 HI, approximately, 

1.256 

where I is length in centimeters of the particular portion being cal- 
culated. The total ampere-turns will tken be the sum of the several 

ni 

partial ampere-turns. See also tables on page 298 for -=- • 



I48 ELECTRICAL AND MAGNETIC CALCULATIONS. 



IX. 

THE E.M.F. OF DYNAMOS AND MOTORS. 

44. Bipolar, Direct Current Machines. — There are 
three elements on which the electromotive force of a 
dynamo machine depends ; namely, the magnetic flux 
through the armature, the number of armature conductors 
and the speed of rotation of the armature. All this may 
be summed up by saying the electromotive force depends on 
the rate of change of the magnetic flux through the armature 
loops. 

Representing the flux in maxwells by <£, the number of 
armature conductors counted all the way around by n, the 
speed in revolutions per second by v, the E.M.F. set up 
in an armature will be, in C.G.S. units, 

E = <f>nv. (78) 

Or expressing E in volts the E.M.F. is 

* = —> r (19) 

Evidently any one of the four quantities in this formula 
may be found when the other three are given, or when 
conditions are given making it possible to determine the 
other three. Thus 

io 8 E 10 8 E io 8 E 

<j> = ; n — — - — ; v— — 

nv cf>v <pv 



THE E.M.F. OF DYNAMOS AND MOTORS. 1 49 

Example. — A dynamo, drum armature, has 50 com- 
mutator segments and 50 coils of wire No. 10, two turns 
to each coil, which are rectangular in shape, 8" X 12"; 
the density of flux through the armature which runs at 
2500 r.p.m. is 10,000 gausses. What E.M.F. does it 
generate ? 

Solution. — There are 50 X 2 = 100 turns, and since 
for each coil there are two surface wires, there are 
in this case 100 X 2 =200 wires, or surface conductors. 
v= 2500 -j- 60 = 4if r.p.s.; <£ = 10,000 X 8 X 12 X \ 5 - = 
6,000,000 maxwells approximately. Hence 
_ <bnv 6 X io 6 X 200 X 4it 

E = ^— r- = r ^-2 = COO VOltS.. 

IO 8 IO 8 ° 

Example. — How many coils of wire of 2 turns each 
would it require on this armature to generate 250 volts ? 

Solution. — 

io 8 xE io 8 X2Co 

n = — = = 100 conductors. 

cj>v 6 x 10 x 4if 

Hence there will be ±%$- =50 coils of one turn each or 
25 coils of 2 turns each. It would perhaps be better to 
use 50 coils of one turn each and 50 commutator bars. 

Example. — If we wish to change the dynamo pulley so 
as to make a 200 volt machine of it, the pulley must be 
selected so as to give what speed in r.p.m. ? 

Solution. — 

io 8 X E io 8 X 200 

v = = = -Z7.1 r.p.s. 

<f>n 6x iox 100 °°^ ^ 

Hence the speed of the pulley should be 33^ X 60 = 
2000 r.p.m. 



150 ELECTRICAL AND MAGNETIC CALCULATIONS. 

45. Alternators. — Alternators usually have all the 
armature coils connected in series instead of the two 
halves in parallel as in bi-polar direct current machines. 
The E.M.F. will therefore be double, other conditions 
being the same. Hence a two-pole alternator would give 
an E.M.F. 

*-"-£ : < s °> 

If there are p pairs of poles, the rate of change of 
magnetic lines will be p times as great, and 

2 

As previously stated the average electric pressure is - = 

0.636, while the effective pressure is — — = 0.707. Hence 

V2 

the formula (81) must be increased by the ratio of 

1 2 
these, or by — — -f- - = 1.11. We shall call this value k, 

V2 * 

whence _, 2 <f>nvpk , n . 

In practical machines the value of k will vary somewhat 
from the theoretical value, depending on the width of 
poles, pitch and nature of windings. It generally ranges 
from 1 to 1.2. But there will be on the other hand a 
certain amount of opposing E.M.F. in the opposite sides 
of each coil because of their width, the amount depending 
on the ratio of the width and pitch of poles to the width 
of the coils. This tends to reduce k, so that as an 
average it will not vary much from 1 . 1 1 . 



THE E.M.F. OF D YNAMOS AND MO TORS. I 5 [ 

In alternators multiple wound the electromotive force 
is reduced to that produced only by the number of coils 
in series. If half are in series, then 

*=*=?• (*) 

It is to be observed that n is always the number of con- 
ductors counted all the way round the surface of the 
armature ; and <j> is the maxwells of flux out of one pole 
into the armature. 

We may modify (82) by substituting f, the frequency, 
for the product vp to which it is equal ; and instead of 
k put its value, 1.11. We then have 

E^-^. (84) 

If it is desired to find the flux which each pole must 
carry in order to produce a certain E.M.F. at a given 
frequency when the number of armature conductors is 
fixed, we have from (84) 

<£ = -2- ' (85) 

^ 2.22 nf v 0/ 

For the number of armature conductors, 

E X iq8 /q^ 

n = —, • (85) 

2.22 <f>/ y 

For the frequency from which 7/ may then be obtained, 

Example. — Suppose the conductors of the two-pole 
machine represented in the first, example under section 



152 ELECTRICAL AND MAGNETIC CALCULATIONS. 

50 were all in series and the opposite ends connected to 
two rings so as to supply an alternating current. What 
would be the E.M.F. ? 

Solution. — 

2 d>nv 2 X (6x io 6 ) X 200 X 41I- 

E = a = - '—=. 2- = IOOO VOltS. 

IO 8 IO 8 

Example. — Determine the electromotive force of a 
6-pole alternator running at 2100 r.p.m., the flux being 
24 x io 5 maxwells and there being 200 wires on the 
armature. 

Solution. — 

2 $nvpk _ 2 X 24 X io 5 X 200 X 35 X 3 X 1.1 1 
io 8 io 8 

= 1 1 18.8 volts. 

Example. — What must be the number of maxwells of 
flux furnished by each pole of a machine similar to the 
last for 560 volts, and what cross section of poles ? Also 
what width of poles to be equal to one-half the pitch? 

Solution. — 

Ex io 8 560 X io 8 „ 

d> = = =1,201,200 maxwells, 

2.22 nf 2.22 x 200 X 3 X 35 

For dynamo poles B may be, say 12,000 gausses. 

Hence 

<f> 1,201,200 

A = -£ = — = 100 sq. cm. 

B 12,000 

The circumference of the pole circle is very nearly 
80 cms. Since there are 6 poles and the distance apart 
is equal to the width, each must be8o-j- 12 = 6.66 cms. 



THE E.M.F. OF DYNAMOS AND MOTORS. 1 53 

The length of pole face and armature core must then be 
ioo-i- 6.66 =15 centimeters. 

46. Multipolar Direct Current Dynamos. — The usual 
method of winding Gramme ring armatures is to wind 
the different coils the same way continuously around the 
ring. Now when the coils are joined together, the end of 
the first may be joined to the beginning of the second, 
the end of the second to the beginning of the third, etc. 
There will then be as many brushes and as many arma- 
ture circuits as poles. This is called the multiple wind- 
ing. Thus a 4-pole machine requires 4 brushes — two 
pairs at right angles, and 4 circuits in the armature all in 
parallel. Evidently the brushes of like polarity may be 
connected together, thus putting all the current in a single 
external circuit. Also the coils may be so cross connected 
at the commutator end of the armature as to require but 
two brushes set at an angular distance apart equal to the 
pole pitch. The E.M.F. of such an armature is 

*-£■ .« 

As before, <£ is the maxwells of armature flux from each 
pole, n the number of armature conductors counted all 
the way around, and v the speed in r.p.s. 

On the other hand the coils may be joined by connect- 
ing the end of the first to the end of the second, the 
beginning of the second to the beginning of the third, 
the end of the third to the end of the fourth, the begin- 
ning of the fourth to the beginning of the next, etc. In 
a 4-pole and 4-coil machine, the beginning of the fourth 



154 ELECTRICAL AND MAGNETIC CALCULATIONS. 

is connected to the beginning of the first, thus completing 
the closed circuit. This is called the series winding. It 
requires two brushes and has two circuits through the 
armature in parallel. If there are p pairs of poles the 
E.M.F. for the series winding is 

In both cases there are as many parts on the commutator 
as coils, and a connection to the commutator is made at 
each junction of two coils. 

For drum armatures the multiple winding is made by 
putting each coil on a chord so as to cover an angular 
space on the armature about equal to the pitch of the 
poles. The coils overlap each other, and the winding is 
done exactly as for 2-pole machines. 

The series winding is made by winding forward along 
the drum, back at an angular distance from first wire 
about equal to the pitch, forward at the same angle from 
last, and so on till the surface of the armature has been 
turned once over ; if several turns are to be put in each 
coil they must all be wound successively in the same slots. 
When one coil is done, the next is put on in its proper 
slots in the same way as the first, etc., until all are on. 
Then the end of the first is joined to the beginning 
of the next, etc. This arrangement requires but two 
brushes. The winding, if spread out, is a wave, like 

,his: J^ru^_n_ 

Hence the name, wave winding. 



THE E.M.F. OF DYNAMOS AND MOTORS. 155 

The same formulae (&&) and (89) apply to the mul- 
tiple and series, respectively. 

When the number of pairs of poles is even, the two 
brushes in the series winding will be at an angular dis- 
tance apart equal to the pole pitch. If the number of 
pairs be odd, however, the brushes will be diametrically 
opposite. 

Example. — Two drum armatures for 6-pole machines 
are exactly alike, except one is multiple wound, while the 
other is series. The following constants are known : 
Flux 6 X io 6 maxwells ; surface conductors 500 ; speed 20 
r.p.s. Determine their respective E.M.F.'s. 

Solution. — For the multiple wound, 

^ <bnv 6 X io 6 X coo X20 , 
£ = ?— = g = 600 volts. 

IO 8 IO 8 

For the series wound, 

_ <bnvp 6Xio 6 X cooX 20X ^ 
E = ?—?- = ?—, 3 = 1800 volts. 

IO IO 8 

Example. — What will be their respective resistances 
and current capacities ? 

Solution. — If wound with wire of the same size the 
multiple wound can carry 3 times as much current as the 
series wound, because with 6 brushes and 6 circuits in 
parallel, each circuit carries 1 of the total output in 
amperes. On the other hand, the series wound armature 
has 2 circuits in parallel, each carrying \ of the total 
current in amperes. Hence the former can carry 3 times 



I56 ELECTRICAL AND MAGNETIC CALCULATIONS. 

as much as the latter. The output in watts is the same in 
both cases. 

The resistance of the multiple wound will be, for the 
reasons just stated, \ of that of the series winding. The 
latter has \ the resistance of all its armature wire, since 
two halves of it are in parallel. The former has 3^ 
the resistance of all its wire, since its 6 sixths are in 
parallel. 

Example. — A machine is required to give 560 volts 
and to run at a speed of 1800 r.p.m. Suppose it is built 
with 6 poles, what area must each pole face present to 
the air gap, not allowing for leakage ? 

Solution. — From (89) 

E x io 8 

<f> = . 

nvp 

Assuming 400 armature conductors series wound, 

560 X io 8 _ „ 

<£> = — = ic.cc x io 6 maxwells. 

400 X 30 x 3 ° o:> 



then A = -±-±z . = 1000 sq. cms. 



Allow B = 15,550 gausses, 

15-55 X io f 

15.55 X 1Qi 

Example. — What will be the diameter of the armature 
if the pole face may be taken 2\ times as long as wide, 
the pole width being equal to one-half the pitch ? 

Solution. — The width of the pole face is 



Viooo -5- 2.5 = 2 ° cm « 
The length of the face is 

1000 -7- 20 = 50 cm. 



THE E.M.F. OF DYNAMOS AND MOTORS. 157 



This is also the length of the armature core. The cir- 
cumference of the pole circle is therefore 20 x 2 x 6 = 



240 cm 




Fig. 10 (a). 



-s- 3.1416 = 76.4 cm. 
Allowing an air gap 
of 0.2 cm., the diam- 
eter of the armature 
must be 76 centi- 
meters. 

47. Polyphase Ar- 
matures. — Armatures 
wound two-phase are 
usually connected in 
one of two ways. 
First, the two wind- 



ings which are put on the core ninety electrical degrees 
apart are connected independently each to its pair of 
rings as shown in Fig. 10. 



j wvmv^ 



This requires four 
wires for its external 
circuits. The armature 
winding may be slightly 
modified by joining the 
circuits at their middle 
points ; that is, at the 
points where they cross 
over in the figure. 

Second, the finishing 
ends of each set of 
windings may be joined 
together, thus requiring but three wires for the external 



Fig. 10(b), 



158 ELECTRICAL AND MAGNETIC CALCULATIONS. 

circuit. See Fig. 11. In the first case, that of inde- 
pendent external circuits, whether the coils are con- 




Fig, n (a). 

nected at the center or not, the voltage of each of the 
circuits A and B will be 




E = 



or 



2 $\nvpk 



io c 



n<t>nf 



io ( 



(9o) 



(9i) 



In these formulae n 
is the number of con- 
ductors on the arma- 
Fi «- IT <*>• ture surface in both 

sets of coils, /= vp is the frequency, and <£ is the flux 
from each pole. It is obvious that each armature cir- 
cuit constitutes an independent alternator, and hence the 
same formula as for plain alternators applies, (82) and 
(84), if we count n the number of surface conductors in 
each phase. 

In the second case, Fig. 11, the E.M.F. between either 



THE E.M.F. OF DYNAMOS AND MOTORS. 1 59 



A or B and the common line C, that is, the machine E.M.F. 
will be 



n 



E = 



2.2 2(f)- Vp 

2 1.1 1 <$>nf 



10 10" 

This is the same as the first case. But between A and B 
the E.M.F. is the resultant of two E.M.F.'s at right angles, 
or 90 degrees apart; hence 

£'= V2E, 

which substituted in (90) gives 

/- , n 

V2 X 2.22 d>-7Jfi , . 

e = . 2 = l -^t nf - (92) 



io l 



IO l 




Fig. 12. 

If the current in each circuit as measured by an alter- 
nating current ammeter placed in the wire A or B be /, 
then the current in the common wire C is the geometric 
sum, or resultant of the two equal currents at right angles ; 
namely /' = V2 X L This assumes the circuits to be 
properly balanced. 



l6o ELECTRICAL AND MAGNETIC CALCULATIONS. 

Armatures wound three-phase are also usually connected 
in one of two ways. First, the coils for each of the 
three phases are put on 120 electrical degrees apart, 
and connected as shown in Fig. 12, 1 to 2, 2 to 3, 
1 to 3. 

This is the mesh arrangement, requires 3 rings, 3 wires 
for the external circuit, and gives an E.M.F. between any 
two wires which is equal to the E.M.F. set up in each 
phase. This is 

2.22 9-/ 

This may be written 

0.74 <£/*/* 



£ = 



io c 



n is the number of surface conductors counted in all the 
phases, and f = vp, as before. The current in any line 
wire is V3 times the current in any one-phase winding 
in the armature. If /' be the current in coil 1, the cur- 
rent measured by an amperemeter in any line will be 

f=^f'=I. 73 2l'. (94) 

This relation comes from the fact that the line current is 
the resultant of two equal armature currents 120 degrees 
apart ; that is, it is the third side of an isosceles triangle 
opposite the angle of 120 . Hence it is V3 times each 
of the two equal sides. 

Second, one end of each winding is connected to a 
common junction, the other ends being joined respec- 



THE E.M.F. OF DYNAMOS AND MOTORS. l6l 



tively to one of the three rings according to the figure 
below, Fig. 13. 





Fig. 13. 

This is the star arrangement, and the E.M.F. of the 
machine, that is, the E.M.F. between any two rings, is 
the geometric sum of the two E.M.F.'s at 120 apart in 
phase. Hence 

V3 X 2.22 X cf>X \nf 



E = 



Reducing, 



E 



io° 
1.28 <f>nf 
7tf~ 



(95) 
(96) 



n is the whole number of surface conductors in all the 
phases. If E r be the E.M.F. in any one of the three 
armature windings, then the machine E.M.F. is 

E=\[^XE'=i. 73 2E'. (97) 

The current in any wire of the circuit is the same as the 
current in the armature coils of the corresponding phase. 

Armatures of the monocyclic type which are virtually 
polyphase when used for running motors are wound just 
as any ordinary single-phase alternator, that is, with a 
continuous winding, the ends being connected each to a 



1 62 ELECTRICAL AND MAGNETIC CALCULATIONS. 

ring ; but in addition another teaser winding is put on 
as follows : one end is connected to the middle of the 
main winding, and its coils, each having one-fourth as many- 
turns as those of the main 
winding, are placed 90 elec- 
trical degrees from them. 
The other end is then con- 
nected to a third ring placed 
between the other two. For 
lighting only the main wind- 
ing and the outside rings are used. For power the three 
rings and three circuit wires are used. The windings are 
related as in the diagram, Fig. 14. 





Fig. 15. 

Between A and B is the main machine voltage supplied 
to transformers for the lighting circuits. When induction 
motors are to be supplied, a third wire from ring C and 
two transformers are required. Fig. 15 represents the 
method of connecting up the transformers for power pur- 
poses on monocyclic circuits. 

The E.M.F. for lighting, that is, between A and B, is 



E = 



2 $nvpk 



io c 



(98) 



THE E.M.F. OF DYNAMOS AND MOTORS. 1 63 

The E.M.F. between A and C or B and C is the re- 
sultant of two E.M.F. 's at right angles, one being \ of the 
other; or one is \ of the total voltage, the other \. 
Hence between the middle ring and either outside ring 
the E.M.F. is 

V(i) 2 + (i) 2 = 0.56 of the main E.M.F. 

Therefore the formula for this is 

1. 12 <f>?izpk , s 

£ = —TJ (99) 

n is the number of surface conductors in the main 
winding. 

If the main E.M.F. is 1040 volts, the E.M.F. between 
the middle ring and either outside one is 0.56 x 1040 = 580 
volts. It is recommended that the transformers for the 
induction motors be 9 to 1 or 4 to 1 , instead of 1 o to 1 or 
5 to 1, so as to give a higher voltage at the motor termi- 
nals to allow for the inductive drop in the motors. If the 
transformers are connected as indicated, the three second- 
ary voltages are approximately equal. 

Example. — A two-phase ring armature with independent 
circuits is to be constructed for 520 volts at a frequency 
of 125. How many turns of wire are necessary for each 
phase when the flux from each pole is io 6 maxwells? 

1 . 1 1 <f>nf 



Solution. — Since E = 



io 8 



£X io 8 C20 X io 8 P 

n = — = ^ = ^72 surface wires. 

I.IKjA I.HXIO b Xl25 

Since each phase contains one-half of the total, each will 
require 1 of 372 =186 conductors. 



1 64 ELECTRICAL AND MAGNETIC CALCULATIONS 

Example. — Suppose the same voltage is required in a 
two-phase machine for three-wire circuits. How many con- 
ductors would be necessary ? 

Solution. — Since the machine voltage, that is, the 
E.M.F. between either outside wire and the common 
middle wire, must be 520 volts, the same as before, each 
phase winding must give 520 volts, as before. Hence 
the same number of turns is required as in the last 
example. But the voltage measured between the two 
outside mains will be 

V S20 x 2 = ^J/ = 73s volts. 

Example. — Let it be required to wind the armature 
for three-phase currents, delta, or mesh winding. How 
many turns will be required in each phase ? Also for a 
line current of 20 amperes primary, how many amperes 
will flow in the armature winding? 

Solution. — For 3 -phase mesh 

0.74^/z/ m 



whence 



10 8 



jEXio 8 520X10 8 

n = J> = e = 55 8 - 

0.74 4>f o.74xio 6 xi25 

Therefore each phase requires i of 558 = 186 wires. 
Also armature current = line current -*■ V3. Then 

_, I 20 
I = — — = ==11.5 amperes. 

V3 I -73 2 



THE E.M.F. OF DYNAMOS AND MOTORS. 1 65 

Example. — A similar machine is to be wound 3 -phase, 
star connected. How many turns are required in each 
phase ? 

Solution. — Referring to the figure given, it is appar- 
ent that each phase must generate the required line 

E.M.F. ■*- Vi ; or E' = —-= = -^- = 300 volts. 

V3 !-732 

Hence 300 Xio 8 

# = — ~ « = 3 2 4, 

0.74X io 6 X 125 

and ^ n = 108 wires in each phase. 

Example. — How many turns must be put in the main 
winding of a monocyclic generator for an E.M.F. of 2080 
volts, assuming </> to be 2 x io 6 maxwells andy= 125 
cycles ; also how many in the teaser winding so that the 
primary E.M.F. between the outside mains and the teaser 
line shall be 1100 volts ? 

Solution. — From (98) the main E.M.F. is 

2.22 <f>?if 

& = -„ — ; 

IO 8 

hence ^Xio 8 . 2080X10 8 

n = — = = ■ — = = 374. 

2.22 4>f 4.44 x io b x 125 

Also, since the primary voltage for motors is to be 1100, 
the E.M.F. for teaser coils must be 

J2 t =y noo 2 -f^-^j = 358 volts 

Therefore 

358 X io 8 



2.22 X 2 X 10 X 125 



= 64 wires. 



1 66 ELECTRICAL AND MAGNETIC CALCULATIONS. 

48. Original Problems. — 1. What flux will a two-pole 
direct current drum type machine be required to supply 
for 220-volt lamps, say a 240-volt machine running at 1500 
r.p.m. when there are to be 100 coils of 2 turns each on 
the armature? = 2.4 x io 6 maxwells. 

2. The following data are taken from a small Westing- 
house bipolar incandescent dynamo. Armature body 8^- 
inches x 8 inches, 100 grooves, two wires per groove. 
Normal speed 2100 r.p.m. It is also determined that the 
armature flux is 1,671,000 maxwells at full load. What 
should be the voltage of the machine under load when 
running at normal speed ? E = 117 volts. 

3. A T.-H. incandescent dynamo, motor type 3 D, 
capacity 3000 watts, furnishes the following data. There 
are 64 coils on the armature, 3 turns per coil ; speed 
2400 r.p.m., E.M.F. at 20J amperes, 116 volts. What 
should be the armature flux under the conditions of load 
named ? <£= 755,208 maxwells. 

4. How many maxwells are due to the compounding 
when the above machine tests no volts on open circuit? 

Approximately, <f) c = 39,062 maxwells. 

5. The pulley on the machine mentioned in problem 2 
has a diameter of 6 inches. At what speed must the 
machine run and how large a pulley will be required to 
give 120 volts, when the engine to which it is belted has 
a pulley 36 inches in diameter running at 350 r.p.m. ? 

Speed = 2154 r.p.m. 
Pulley = 5.85" diameter. 



THE E.M.F. OF DYNAMOS AND MOTORS. 1 67 

6. Suppose the armature in problem 3 is to be rewound 
for 125 volts, using the same fields but changing the com- 
mutator if necessary and running at the same speed as 
before. How many turns per coil will be necessary ? 

n = 414 wires = 208 turns. 
Hence use 52 segments on commu- 
tator, and wind 4 turns to each coil. 

7. The following data are known about a certain 
Westinghouse pony alternator. Speed 2000 r.p.m. 
E.M.F. 50 volts. There are 8 poles and 6 turns per 
coil on the armature. If we may assume the average 
constants to apply, what must be the flux from each 
pole? <£ = 176,000 maxwells. 

8. An 8-pole alternator is rated at 1500 r.p.m. There 
are 12 turns per coil. If an estimate places the flux at 
2,440,000 maxwells, what voltage should the machine give ? 

E = 1040 volts. 

9. What will be the size of armature in the machine 
in problem 8 if y 1 ^ inch is allowed for clearance, and the 
pole width is one half the pitch, not considering any 
leakage, assuming B = 10,000 gausses ? 

Length = 9" = 22.8 cms. 

Pole width = 4.2"= 10.7 cms. 

Pitch circle = (8 x 10.7) X 2 = 17 1.2 cms. 

Diameter = 171.2 -r- 7r = 54.5 cms. 

Armature diameter = 54 cms. 

1 o. How many poles will be required in an alternator for 
a voltage of 1040, surface conductors 200, flux 2,342,000 
maxwells, speed 1200? Find frequency, p = 5 pairs* 

/= 100. 



1 68 ELECTRICAL AND MAGNETIC CALCULATIONS. 

ii. What flux will be required by a drum armature 
wound for 6 poles, direct current, 240 volts, 1200 r.p.m., 
multiple connected, n = 360 ? cj> = 3,333,000 maxwells. 

12. What E.M.F. would this last machine give if the 
armature circuits were series connected ? 

E = 720 volts. 

13. A 4-pole direct current dynamo is built for 120 
volts under load. How many wires must be connected 
on the drum for a speed of 1200 r.p.m., multiple circuit 
armature, flux 3 x io 6 maxwells ? n = 200 conductors. 

14. A half horse-power motor, 4 poles, is to be built 
to run at 2400 r.p.m., series connected. It is calculated 
that the polar faces are to be i ;/ x 4". The E.M.F. is 
to be 112 volts. A density of flux in the air gaps is 
assumed to be 10,000 gausses. How many turns per 
coil must be wound on the drum, for a 24-coil armature ? 

n = 560 wires = 24 per coil. 

15. What flux must be sent into a 2-phase armature 
having 200 wires in each phase for an E.M.F. of 1040 
volts, the machine being 12-pole and running at 900 
r.p.m., connected for 4-wire circuits ? 

<£ = 2.6 x io 6 maxwells. 

16. What would be the required magnetization for a 
voltage of 520 when the windings are connected for a two- 
phase, 3 -wire circuit, other conditions the same as in 15 ? 
Also what is the frequency? What speed would give 
a frequency of 120? <£ = 1.3 x io 6 maxwells. 

/=9°- 

v = 1200 r.p.m. for 120 cycles. 



THE E.M.F. OF DYNAMOS AND MOTORS. 1 69 

17. How many turns must be wound in each phase of 
a two-phase alternator having ten poles, for 2080 volts, 
for a frequency of 60 and an estimated flux of 12 x io 6 
maxwells, circuits to be 3-wire ? What is the speed ? 

n, each phase, = 130. 
Speed = 720 r.p.m. 

18. How much current must each phase winding be 
capable of carrying in problems 16 and 17, when it is 
desired that the wattmeter in the circuit read 50 K.W. 
full load in each case ? 

Effective line current in 16 = 96 amperes. 
Total =192. 

Effective line current in 17 = 24 amperes. 
Total = 48. 

19. What is the E.M.F. between the outside wires in 
the machine described in problem 17? 

E = 2941.5 volts. 

20. What should be the E.M.F. of the following 3- 
phase, star connected alternator? Normal speed 1000 
r.p.m., 12 poles, 560 conductors each phase, flux from 
each pole 804,000 maxwells. Also what is the frequency ? 

E.M.F. = 1729 volts. 
f= 100. 

21. A 3-phase mesh connected alternator having ten 
poles, running at 720 r.p.m., flux 7,600,000 maxwells, has 
a capacity of 250 K.W. What is the brush E.M.F., and 
the current in each phase winding, the total surface con- 
ductors being 600 ? E = 2025 volts. 

/= 41 amperes per phase. 



\yo ELECTRICAL AND MAGNETIC CALCULATIONS. 

22. Let the armature in 21 be star connected. How 
will the E.M.F. and current differ from those just 
obtained? E.M.F. = 3502 volts. 

/= 41 amperes per phase. 

23. Find the flux and the current per coil for a 3- 
phase machine, first when star connected, second when 
mesh connected, having given the following data. Sur- 
face conductors 864, poles 8, speed 900 r.p.m., brush 
E.M.F. 2500 volts, capacity 120 K.W. 

Star, cj> = 3.7 x 1 o 6 maxwells. 

/= 28 amperes. 
Mesh, <f> = 6.5 x 10 6 maxwells. 
/= 16 amperes. 

24. How many turns must be put in the main winding 
of a monocyclic generator, /= 60, p — 5, <j> = 5 x 10 6 , 
E = 1080, capacity =150 K.W. ? Also if the teaser 
winding has just \ as many turns, how many volts should 
be measured between the main and teaser line wires ? 
What is the effective current and the speed ? 

// = 10 coils, 8 turns each. 
E.M.F. for motors = 605 volts. 
/= 140 amp. ; speed 720 r.p.m. 

25. What flux must be provided for a monocyclic 
machine for 2160 volts, 10 poles, /= 60, main winding 
= 200 conductors? What is the current capacity of the 
wire when the machine is rated at 200 K.W? Also 
what should be the speed? <f> = 8,108,100 maxwells. 

/= 92.6 amperes. 
v == 720 r.p.m. 



CALCULATION OF FIELDS. I J I 



X. 

CALCULATION OF FIELDS. 

49. Two-Pole Continuous Current Machines. — We are 
now to apply to dynamos the principles of magnetic cir- 
cuits which were worked out in a previous chapter. 
However, there are some practical details in connection 
with dynamo fields which it will be well to keep in mind 
in determining magnetomotive forces. The following 
examples will show that there are several parts in the 
magnetic circuit, — field cores, yoke, two air gaps, and 
armature core. Account must be taken of the flux den- 
sity, length and cross section of each part. From these 
reluctances are found which multiplied by the total flux in 
each part gives the M.M.F. required for each portion of 
the circuit. 

Example. — A bipolar, drum armature, direct current 
machine requires 6 x io 6 maxwells of flux. The armature 
core is 8" x 12". The air gaps including clearance and 
copper and insulation are each 1 centimeter in length. 
The poles are separated above and below the armature 
4 inches. The field cores are 11" in diameter and 15" 
long. Polar heads, that is, the portion above the wind- 
ing, average 8 inches in length of flux. The yoke 
between core centers is 20" and has a cross section of 
100 square inches. The leakage coefficient is assumed 
to be 1.3. Determine the separate reluctances, the 



172 ELECTRICAL AND MAGNETIC CALCULATIONS. 



M.M.F. in gilberts for each, the ampere-turns, and depth 
of winding and length of wire for an exciting current of 
10 amperes. The armature is Norway iron, the rest is 
steel. See Fig. i6. # Check results by curves, p. 229. 



Armature 






1 — * — 1 


-TO*- 


* 1 

1 






1 

! ! 




it 






19 

CO 

1 
1 

1 




<-- J 27.6— * 
1 

CO 

1 






---50--- 


- ip 

1 



Fig. 16. 

Diameter, /= 8" = 20 cm. 
Area, A = 20 x 30 = 600 sq. cm. 
Induction, B a = 6 x io 6 ^ 600 
= 10,000 gausses. 

O.OOOI 



Reluctivity, k = 



1 — 0.000059 X io 4 
= 0.000244 oersted per c c. 
20 



Reluctance, (R = 0.000244 x 

600 

= 0.000008 oersted. 

M.M.F. = 6 x io 6 x 0.000008 = 48 

gilberts. 

48 



Ampere-turns, nl — 

1.256 

* See also Example 4, p. 222. 



= 38. 



CALCULATION OF FIELDS. 1 73 

Air gaps : Length, / = 2 cm. 

[22 71-— (iO X 2)1 X ^O 

Area, A = = * JJ ^- 

2 

=3 736.72 sq. cm. 

Reluctivity, k = 1 . 

2 
Reluctance, (ft = 1 X — 7 — = 0.00271 

736.72 

oersted. 

M.M.F. = 6 x io 6 x 0.00271 
= 16,260 gilberts. 

Ampere-turns, nl ' = 16,260 -5- 1.256 
= 13,000. 

Poles : Length, / = 20 cm. 

Area, ^ = 736.72 sq. cm., — same as air 
gap. 

Induction, B p = 6 x io 6 -r- 736.72 =8,144 

gausses. 

^ , . . , 0.00041; 

Reluctivity, k = -^— 

1 — 0.00005 1 X 0,144 

= 0.00077 oersted per cc. 

_ • _ 20 

Reluctance, 61 = 0.00077 x — 7 — X 2 

736.72 

= 0.000418 oersted. 

M.M.F. = io 6 x 6 x 0.0000418 
= 250 gilberts. 

Ampere-turns, nl = 2.^0 ~- 1.256 = 200, 



174 ELECTRICAL AND MAGNETIC CALCULATIONS, 

Cores: Length, /= 15" = 37.5 cm. 

2 

Area, A = 27.6 X 0.7854 = 600 sq. cm. 

Induction, B c = (6 X io 6 x 1.3) ■+■ 600 
= 13,000 gausses. 

Reluctivity, k = 0.00045 ft 

1— 0.000051 x 13 X io 8 

= 0.00133 oersted per cc. 

Reluctance, (ft =0.00133 x 7^ X 2 

600 

= 0.000166 oersted. 

M.M.F. = 6 x io 6 x 1.3 X 0.000166 

= 1295 gilberts. 

Ampere-turns, nl ' = 1295 -5- 1.256 
= 1031. 

Yoke: Length, /= 50 cm. 

Area, A = 100 sq. in. = 645 sq. cm. 
Induction, B y = 6 X io 6 X 1.3 -*- 645 
= 12,093 gausses. 
0.00045 



Reluctivity, k = 



1 — 0.000051 X 12,093 
= 0.00117 oersted per cc. 

Reluctance, (R = 0.00117 X -? — 

645 
= 0.00008 oersted. 

M.M.F. = io 6 X 6 X 1.3 X 0.00008 

= 625 gilberts. 

Ampere-turns, nl = 625 -f- 1.256 

= 500. 

Hence the total ampere-turns = 38 + 13,000 + 200 + 103 1 
4- 500 = 14,769 = 7385 on each core. For 10-ampere 



CALCULATION OF FIELDS. I 75 

current, this requires 7385 -f- 10 = 738 turns on each 
core. About 1000 circular mils per ampere should be 
allowed in field windings. Hence 10,000 circular mils is 
the necessary cross section of the wire, corresponding to 
No. 10 B. & S., whose diameter is 102 mils bare, or say 
120 mils double cotton covered. Therefore allowing £ 
inch at each end of the core for collar and insulation, 
there will be in one layer [(15 — ^) -f- 0.120] =120 turns. 
This requires a depth of winding = 738 -s- 120 = 7 layers, 
nearly, allowing for insulation on cores before wire is put 
on. Only about 6^ are needed for the excitation. 120 x 
6^ = 780 turns to the core. The cores may have been 
shortened slightly, or a little heavier insulation put on 
against the collars so as to require just seven layers. 
Depth of winding =7X0. 120 = 0.84 inch. Total addi- 
tion to the diameter of the field = o.84X 2 = 1.68 inches. 

1.68 _ : 

= A or core diameter. 

11 

The average diameter of the finished field is 11 + .84 

= 11.84 inches. Circumference of the mean turn = 

11.84 X 7r = 37.2 inches. 

Total length of wire necessary for excitation = — ^— 

= 4575 fee *. 

4575 X IO - 8 
R f = 321^ — = 4#7 hms cold. 

10,382 

J? f hot = 4.7 + 20% of 4.7 = 5.64 ohms. 

Wiring table gives for No. 10 wire 1.034 ohms per 1000 ft. 
Hence 

ify (from table) = 1.034 x 4.575 = 4.7 ohms cold. 



176 ELECTRICAL AND MAGNETIC CALCULATIONS. 

50. Consequent Pole Type. — Example. — Let it be 
required to determine the winding of a consequent pole 
field for the armature in the last example, the speed, 
voltage, flux, etc., to remain the same, and the dimensions 
of the field parts to be as follows : diameter of the 
cores (four) 21.5 cm.; average length of poles for flux 
lines 1 8 cm. ; air gap 1 cm. ; separation of pole tips 1 o 
cm.; length of cores 20 cm.; height of yokes 65 cm.; 
total length of machine 90 cm. The coefficient of leak- 
age for this type is taken at 1.7. Check by curves, p. 229. 
Solution. — Since there are two magnetic circuits in 
parallel in this type it will be well to calculate each cir- 
cuit separately, each side furnishing one-half the re- 
quired flux. See Fig. 17. 

Armature: Diameter, /= 20 cm. 

Area, A — 20 X 30 = 600 sq. cm. = 300 

each half. 
Induction, B a — 3 X io 6 -r 300 
= 10,000 gausses. 

^ 1 . . 7 0.0001 

Reluctivity, k = 2 

1 — 0.000059 x IO 

= 0.000244 oersted percc. 

20 

Reluctance, 61 = 0.000244 X 

300 

= 0.000016 oersted. 
M.M.F. = 3 X io 6 X 0.000016 

= 48 gilberts. 
Ampere-turns, nl = 48 -4- 1.256 = 38. 
Air gaps : Length, /= 2 cm. 

Area, A = 736.72 -5- 2 = 368.36 for half 
the flux. 



CALCULATION OF FIELDS. 



17? 



Reluctivity, k = 1. 
Reluctance, (R=iX 



= 0.00542 oersted. 



368.36 

M.M.F. = 3 X io 6 X 0.00542 = 16,260 gilberts. 
Ampere-turns, nl '= 16,260 h- 1.256 == 13,000, 

Poles : Length, / = 18 cm. 

Area, A = 736.72 -5- 2 = 368.36 sq. cm. 
Induction, B p = 3 X io 6 -r- 368.36 
= 8144 gausses. 
0.00045 



Reluctivity, k = 



1 — 0.000051 x 8144 
= 0.00077 oersted per cc. 
...90 



*1-2-.5* 




* 






f 


1 
1 








"?• 


*F 






1 
1 


* — ■ 


— _U-> 

CN 


-\ iji r- 


<- 


\ 


| 


i 
i 

1 

1 

i 

1 
1 
1 


* 




«.— 20— > 


1 

1 

1 
1 

1 

* 

1 
1 

1 


1 

1 


C 




! 

1 
t 






1 
1 




<r 


J S v- 


^ 


1 


\ 

1 
t 
i 
1 






\ 






Fig. 17. 








Reluctance, (R = 0.00077 X —77; — -z 

368.36 




= 0.000038 oersted. 




M.M.F. = 2 X 3 X io 6 X 0.000038 




= 228 gilberts. 




Am 


pere-turn 


S, /z/ = 22 


8 -T- I.2C 


;6 = 


l82 



178 ELECTRICAL AND MAGNETIC CALCULATIONS. 

Cores : Length, / = 20 cm. 

Area, A = 21.5 X 0.7854 = 363 sq. cm. 

Induction, B c — 3 X io 6 X 1.7 -5- 363 

= 14,060 gausses. 

^ , . . , 0.0004c 

Reluctivity, k = — — 

1—0.000051X14,000 

= 0.00159 oersted per cc. 

_ , _. 20 

Reluctance, (R = o.ooico X -7- X 2 

= 0.0001752 oersted. 
M.M.F. = 3 X io 6 X 1.7 X 0.0001752 

= 894 gilberts. 
Ampere-turns, nl = 894 -*- 1.256 = 712. 

Yoke : Length, / = 42 cm. for lines of force. 

Area, A = \2\ X 30 = 375 sq. cm. 

Induction, B y = 3 X io 6 X 1.7 -s- 375 

= 13,600 gausses. 

-r. 1 • • t 0.0004c 

Reluctivity, k ~ 



1 — 0.000051 x 13,600 
0.00115 oersted. 
42 



Reluctance, (R = 0.00 1 1 c X 

3 375 
= 0.000128 oersted. 

M.M.F. = 3 x io 6 X 1.7 X 0.000128 

= 652 gilberts. 

Ampere-turns, nl = 652 -5- 1.256 == 519. 

Hence the total ampere-turns for each circuit = ^8 

+ 13,000 + 182 + 712 + 519 = 14,451. Total reluctance 

two paths in parallel, 0.005777 -5- 2 = 0.002888 oersted. 

The total M.M.F. for each circuit = 18,151 gilberts. 



CALCULATION OF FIELDS. I 79 

The machine will require on each of the four cores \ of 
14,451 =7226 ampere-turns. If we connect all the field 
coils in series, the wire will be of the same size as in last 
example, namely No. 10 B. & S. = 120 mils d.c.c. Al- 
lowing 1 centimeter for collars and insulation on each 
core, we have 19 cm. for wire = 7.5 inches, nearly. In 
one layer there will be 7.5 -s- 0.120 = 62 turns which will 
require 722.6 -5- 62 = 12 layers approximately, making a 

i-44 
8.6 



depth of winding 0.120 X 12 = 1.44 inches = 2 x 



= \ of core diameter. 

Mean length of one turn = — — — = 2.7 feet. 

Total length of No. 10 wire for fields = (2.7 x 744) X 
4 = 8035 feet. Total weight (32 lbs. per 1000 feet, ap- 
proximately) = 8.035 * 3 2 = 2 57 ^s. 
Resistance of field winding 

8035 X 10.8 , 

R f = — — — - = S.z ohms cold ; 

8.3 + 20% of 8.^ = 9.96 ohms hot. 
Drop in field winding = 9.96 X 10 = 99.6 volts. 

Suppose now we wind the two halves for 5 amperes 
each and connect them in parallel. Each spool will re- 
quire 7226 -7- 5 = 1445 turns of 5000 circular mils each 
cross section. This corresponds to No. 14 B. & S. = 64 
mils bare, or about 80 mils d.c.c. In one layer there 
will be 100 turns, and 14 layers deep = 14 X 0.08 = 1.12 
inches = \ of core diameter. Mean length of turn = 

(8.6+I.I2)7T . . 

i — '— = 2.5 feet, requiring 1445 x 2.5 = 3612.5 

1 — 



l8o ELECTRICAL AND MAGNETIC CALCULATIONS. 

feet on each core, or 14,450 feet total length, having a re- 
sistance hot of 38 ohms. But since two halves are in 
parallel the field resistance will be % of 38 = 9.5 ohms and 
the drop = 9.5 X 10 =95 volts. The weight of No. 14 
wire will be 14.45 X 13 = 190 lbs., provided the assump- 
tions about insulation are correct. 

The radiating surface for the last case will be (8.6 + 
2.24) 7r X 8 = 272 sq. in. on each coil, or 1088 sq. in. 
total. Total watts lost =95X10 = 950 watts, making 
about 1 watt per square inch. 

51. The Multipolar Type. — The preceding armature 
core is to be wound for a four-pole field and series con- 
nected, thus requiring 3 X io 6 lines of force from each 
pole, instead of 6 X io 6 . The following are the field 
dimensions : mean circumference of pole ring 190 cm. ; 
width 6.5 cm.; length of cores 16 cm. ; width of cores 
8.6 cm.; coefficient of leakage for this type 1.3. The 
field ring and poles are steel. Form shown in Fig. 18. 

Armature : Length (for flux), /= 20 cm. 

Area, A = 10 X 30 = 300 sq. cm. 

3 X io 6 
Induction, B a = *- 300 

2 



Reluctivity, k = 



5000 gausses. 
0.0001 



1 — 0.000059 X 5000 

= 0.00014 oersted percc. 

20 

Reluctance, (R = 0.00014 X 

300 

= 0.0000093 oersted. 



CALCULATION OF FIELDS. 



lSl 



i y to 6 
M.M.F. = ^ X o. 



0000093 



= 13-95 gilberts. 

nl — 13.95 -*- 1.256 = 11 ampere-turns. 

Air gaps : Length, / = 2 cm. 

. 8.6 x 30 
Area, ^4 = — = 129 sq. cm. 



Reluctance, (R = 1 X = 0.0 1 c c oersted. 

129 




Fig. 18. 



/fc&r : 



M.M.F. = 3 X IO x 0.0155 

= 23,250 gilberts. 
nl = 23,250 -r- 1.256 = 18,500 amp.-turns. 
Length, / = 1 6 cm. 

Area, A = — X 30 = 129 sq. cm. 



Induction, B = 



3 X io< 



X 1.3-5- 129 



= 15,120 gausses. 



1 82 ELECTRICAL AND MAGNETIC CALCULATIONS. 

Reluctivity, k = ^2 

i — 0.000051 x 15,120 

= 0.0016 oersted per cc. 

T f\ 

Reluctance, (R = 0.0016 X X 2 

129 

= 0.0004 oersted. 

3 X io 6 
M.M.F. = X 1.3 X 0.0004 

= 780 gilberts. 
nl ' = 780 -*- 1.256 = 621 ampere-turns. 

Yoke : Length, /= 190 -r- 4 = 47.5 cm. 

Area, A = 6.5 X 30 = 195.0 sq. cm. 

t j 4.- z? 3 X io 6 X 1.3 
Induction, B y = -f- 195 



Reluctivity, k = 



10,000 gausses. 
0.00045 



1 — 0.000051 x 10,000 
= 0.0009 oersted per cc. 

Reluctance, (R == o.oooo X — ^- 

J 9S 
= 0.00022 oersted. 

M.M.F. = 3 X Iq6 X Im3 x 0.00022 
2 

= 429 gilberts. 

nl= 429 -f- 1.256 = 341 ampere-turns. 

As before, the coefficient of leakage is applied to the 
yoke as well as to the cores, since it is uncertain just what 
the leakage paths are and our error in making the leakage 
factor apply to the yoke will be on the safe side. 



CALCULATION OF FIELDS. 183 

From our data the reluctance of a single magnetic cir- 
cuit will be 

(R = 0.0000093 + 0.0155 + 0.0004 + 0.00022 
= 0.0 161 oersted. 

Hence the combined reluctance of two parallel paths will be 

0.0161 -f- 2 = 0.00806 oersted. 

Also the combined M.M.F. of two parallel paths will 

be 

M.M.F. = 13.95 + 23,250 + 780 + 429 

= 2 4,473 gilberts. 
nl = 24,473 -T- 1.256 = 19,485 ampere-turns. 

This requires 9742 ampere-turns on each core. If we 
take as exciting current 5 amperes, the number of turns 
will be 9742 -r5 = 1944 on each core, requiring 5000 
circular mils section, corresponding to No. 14 wire = about 
80 mils diameter d.c.c. 

The length of core winding =16 cm. = 6300 mils. 
Hence there will be 6300 -j- 80 = 80 turns in one layer, 
thus requiring 1944 -f- 80 = 24 layers deep = 24 x 0.080 
= 1.9 inches. The mean length per turn = 3^ feet, 
making the total length required on the fields, 

/= 1944 X 3} X 4 = 24,624 feet 

The resistance for the fields connected in series will be 

24.63 X 2.58 = 63.6 ohms cold. 

Since the resistance of copper increases about 1 per cent 
for each 2 C. rise above the air, and allowing 40 rise, 
the resistance will become 

R hot = 63.6 + (1 % of 63.6) X — = 76.3 ohms. 



1 84 ELECTRICAL AND MAGNETIC CALCULATIONS. 

The voltage required for excitation is 

E = 76.3 X s = 381.5 volts. 

If we connect the fields so that two will be excited in 
series and two in parallel, the total resistance becomes 

I Q I X 2 

R = — =19.1 ohms hot. The total exciting cur- 
rent must now be 10 amperes, and the voltage for excita- 
tion becomes 

E = 19. 1 x 10 = 191 volts. 
The weight of wire will be 325 to 350 lbs. 

Fields for alternators and direct current machines of any 
number of poles are determined as in the last example. 

It is impossible to be absolutely exact in these prelimi- 
nary calculations, since it is not definitely known just what 
the leakage paths will be and just what the behavior of 
the particular quality of iron will be under test, also the 
effect of joints, etc. Manufacturers, of course, must 
govern their final design by tests of a finished machine as 
well as by tests of the material used. 

For compound wound machines, if the E.M.F. at no load 
and also at full load are known, or if left for the designer 
to decide, the series winding is determined so as to supply 
flux through the known reluctances to make up the differ- 
ence of these two E.M.F's plus the additional leakage. 
Also if the armature resistance and maximum current are 
given, the armature drop due to the load is E a = IR a . 
Hence if the compounding is only for armature drop, we 
can determine the series winding to supply the extra pres- 
sure necessary. 



CALCULATION OF FIELDS. I 85 

All armature bodies are laminated, so that the actual 
area is reduced 10% to 20%. But since the armature 
reluctance is such a very small part of the total, it will not 
be necessary to correct for this reduction of area due to 
lamination. 

In cases where the general dimensions of a dynamo are 
given, a good plan is to lay down the machine to scale on 

a drawing board, then take from the drawing all distances 

nl 

and areas needed in calculating reluctances or — • 

c 

52. Original Problems. — 1. Determine the shunt wind- 
ing, depth, resistance, etc., for the following machine, two- 
pole, direct current. Armature body 8^" X 8", 100 
grooves, two wires per groove, resistance 0.06 ohm ; 
/= 30 amperes ; speed = 2025 r.p.m. ; E.M.F. no load 
118 volts, full load 120 volts. Field bore %\\ inches; 
core length, single, 6 inches; area 63.68 square inches; 
separation at tips \\ inches ; mean length of lines of force 
in each pole 17 inches ; pole width 8 inches. Assume 
wrought iron throughout. Also calculate the series turns 
for compounding. 

Shunt coil 3350 turns. 

Wire No. 21 = 0.0354" covered. 

R f = 120 ohms. 

Series coil 17 turns of No. 3 = 0.283". 

Resistance series 0.013 ohm. 

2. What will be the required number of series and 
shunt turns for the following two-pole, direct current ma- 
chine ? Armature body 8" X 6" ; 64 coils, 3 turns each ; 
resistance 0.85 ohm ; speed 2400 r.p.m. ; E.M.F. no 



1 86 ELECTRICAL AND MAGNETIC CALCULATIONS. 

load no volts; full load 116 volts. Field cores each 
6 inches long, 5^ inches diameter; pole heads from top 
of cores to center line of armature 15 inches; area of 
pole heads 8 X 3= 24 square inches ; tips 2% inches 
apart. Yoke length for lines 13 inches ; width 8 inches; 
depth \\ inches. Air gap including the wire on the 
armature 0.3 inch. The yoke is of cast iron ; armature 
Norway ; cores and poles dynamo wrought iron. Leak- 
age factor is 1.25. 

Shunt 4850 turns No. 20 wire. 

Series 46 turns No. 8 wire. 

3. A small 50-volt, 8-pole alternator has an armature 
core 5^ inches diameter and 6 inches long, and is wound 
with 8 coils, 6 turns each, No. 14 wire. The external 
diameter of the field ring is 13.5 inches, internal n inches ; 
width 5 inches. The clearance is 0.3 centimeter. The 
poles are 1" X 6", and 2\ inches long. Speed of armature 
2000 r.p.m. All parts are made of wrought iron. Allow- 
ing about 600 circular mils per ampere in fields, find the 
number of turns and depth of wire on each core. 

Field 492 turns, 12 layers of No. 19. 
Total, 8 cores = 3936 turns. 
Depth on each core 0.6 inch. 
R f = 42 ohms if connected in series. 

4. A consequent pole machine similar to the one pre- 
viously worked out has the following data taken from it : 
Armature body 8" X 8" ; slotted for 200 wires, 2 per 
groove; speed 2400; E.M.F. 116 volts; core laminated 
Norway iron ; field cores wrought iron, 6 inches long, 4 



CALCULATION OF FIELDS. 1 8/ 

inches diameter ; yokes cast iron, 3 inches wide, 12 inches 
between core centers ; poles cast iron, 8 inches between 
cores, 5 inches perpendicularly, tips separated 4 inches. 
Air gaps are 0.15 cm. each. Determine the turns neces- 
sary to wind the machine as a shunt motor, allowing 2 
amperes for exciting current. Also calculate the length 
of wire, its depth, resistance and weight. 

11 = 810 turns each core ; 3240 total. 

Length of wire 3825 feet of No. 20. 

R /= 43 ohms, allowing 20% for rheostat. 

Depth with insulation 0.4". 

Weight 12 lbs. 
5. The drum armature for a 25-K.W. machine, 125 
volts, is 1 1.5 inches long, 11.5 inches diameter. B a = 
10,000 gausses; speed 1000 r.p.m. There are 112 arma- 
ture conductors. Fields are Edison type, wrought iron. 
Core length, exclusive of collars, etc., 16.6 inches ; diame- 
ter 1 if inches. Height from top of yoke to lower end of 
poles 42 inches. Total length of yoke 34 inches. Air 
gaps, including winding, insulation, etc., on the armature, 
0.05 inch. Pole tips 3 inches apart. Draw a diagram 
and determine the length of flux paths. Also determine 
the field winding for an exciting current equal to 2.5 per 
cent of output and for a leakage factor for this type 
of 1.6. n = 1750 turns, each core. 

Wire No. 13 ; length 11,060 feet. 

R f — 22.18 ohms. 

Note. — A smaller size of wire may have been used, 

say No. 14 wire, without undue heating. The No. 13 

wire was obtained by taking the resistance which the 

winding at 125 volts must have to allow just the required 



1 88 ELECTRICAL AND MAGNETIC CALCULATIONS, 

exciting current to flow, and assuming at the outset a 
depth of winding about J of core*radius in order to obtain 

the length. Whence d 2 = - — — = 5000 circular mils, 

approximately. This may be used as a satisfactory check 
on any method of determining the field winding, or as 
the primary method, itself, to be checked by others. 

6. The following data for an Edison-Hopkinson ma- 
chine are taken from Thompson's " Dynamo-Electric Ma- 
chinery":* normal output 320 amperes at 105 volts; 
speed of armature 750 r.p.m., diameter of armature 
core 25.4 cm., length 50.8 cm., carrying 40 turns, 2 lay- 
ers deep; wire 1.75 mm. diameter. Magnet cores nearly 
rectangular, rounded corners ; all material wrought iron ; 
length of magnet limb 45.7 cm.; breadth 22.1 cm.; 
width parallel to shaft 44.45 cm * Length of yoke 61.6 
cm. ; width 48.3 cm. ; depth 23.2 cm. Diameter pole 
bore 27.5 cm.; depth of pole piece 25.4 cm.; width 
parallel to shaft 48.3 cm. ; width between pole pieces 
12.7 cm. The area of iron in the armature is 810 sq. cm. 
The angle subtended by the pole face is 129 . Effective 
area of pole is 1600 sq. cm. The air gap is 1.5 cm. De- 
sign the shunt winding for 6 amperes exciting current. 

n = 3260 turns, 11 layers on 
each core, No. 1 2 wire. 
J?/= 17 ohms, approximately. 

7. An 8-pole direct current machine has an armature 
3 1.1 cm. X 13 cm. cross section, Gramme ring type, 0.92 
of area effective, on account of insulation of the ribbon of 

* Dynamo' Electric Machinery , page 413. 



CALCULATION OF FIELDS. 1 89 

armature. Speed is 1722 r.p.m. The armature carries 
768 conductors. B a = 9,500 gausses. External diame- 
ter of armature ring is 147.3 cm - 5 finished, 149.3 cm. 
Internal diameter of ring, 12 1.3 cm.; finished, 119 cm. 
The external diameter of pole ring is 231 cm.; internal 
diameter, 218.4 cm « ; width parallel to shaft, 34.3 cm. ; 
width of cores, 29.5 cm. The poles widen at the air gap 
so the latter is 41 X 34= 1400 sq. cm. The armature 
coils are parallel connected. Determine the E.M.F., the 
reluctance and field winding, the armature being soft iron, 
the fields all cast steel. If the exciting current be 10 am- 
peres, how many turns and how deep on each core ? 

E= 155.7 volts. 

nl = 7333 each spool. 

n = 734 turns each. 

Depth = 1.4". 

(ft/ = 5.214 X 1 o~ 3 oersted. 

8. An iron-clad armature for a multipolar dynamo has 
been designed so that it requires in its teeth from each 
air gap 20.4 x io 6 maxwells of flux. The armature is of 
sheet iron 14 inches long, cross section of teeth under 
pole 151. 5 sq. in. ; length of teeth 1.75 inches; armature 
cross section 124.3 sc l- m - Ah" gap, length 0.25 inch; 
area 538 sq. in. Field poles, cross section 237.2 sq. in. ; 
length 16 inches, cast steel. Yoke, section 132 sq. in.; 
length 28.5 inches, cast steel. One coil on each pole 
will require how many ampere-turns each, leakage coeffi- 
cient being assumed as small as 1.2 ? 

nl = 5850 ampere-turns. 



19O ELECTRICAL AND MAGNETIC CALCULATIONS. 



XL 

ELEMENTS OF DYNAMO DESIGN. 

53. General Considerations. — There is such a diversity 
of elements upon which the judgment must depend in 
the proper design of machines of the same type, to say 
nothing of different types, that only those details which 
illustrate the general principles of dynamo design can be 
given in this chapter. The general theory of the dynamo 
must be understood and the exact character of all the 
material going into its construction. Observation and 
test of various types and sizes of practical machines must 
be made use of. Then the manufacturers must gather 
experience, not only from continuous tests of all material 
used, but from the performance and tests of the completed 
product. In fact, there are so many interdependent 
quantities which themselves depend on the conditions of 
service and exact quality of material, that an approxima- 
tion to exactness is all that can be expected in a pre- 
liminary design. 

54. Useful Data. — (1) The output of dynamos is ap- 
proximately proportional to their weight, or to the cube 
of a linear dimension, if of the same type, since the larger 
must have reduced speed, increased efficiency, greater 
radiating surface, etc. Double the weight of copper and 
double the weight of iron should give double the output. 
The output will be about 6 watts per pound for continu- 



ELEMENTS OF DYNAMO DESIGN 191 

ous current belted dynamos, and 8 watts per pound for 
direct driven multipolar machines. 

(2) The temperature of armatures should generally not 
exceed about 50 C. above the air. Although the subject 
is somewhat indefinite, yet tests tend to show that the 
amount of external surface in an armature should be taken 
so that 1 sq. in. may be allowed for each 2 to 2} watts 
lost in the armature. Of this loss about 1 watt per sq. in. 
will be due to PR loss, and the rest to hysteresis and 
eddy currents. 

For example, if the I 2 R loss is 2% of 100 K.W. = 
2000 watts, the external surface of the armature should be 

= 2000 sq. in. In designing armatures this method 

may be used for determining the size of armature. In 
any event it should be used as a check on any other 
method employed to obtain the required size of armature. 
If armatures are internally ventilated, a smaller area per 
watt can be used. 

The fields should have somewhat larger surface per 
watt lost owing to the fact that the winding is deeper and 
there is no mechanical ventilation as in armatures. Some 
recommend 2 square inches per watt lost in field wind- 
ings. This may be assumed as a safe limit, though in 
many cases only 1 square inch per watt is allowed. 

In very small sizes of machines the temperature is per- 
mitted to go higher, since the same efficiency is not 
attempted as in large sizes. Often as low as 250 to 400 
circular mils per ampere is used for the armature wires 
and 500 to 800 for the field windings. But as a working 
average for ordinary machines 600 circular mils per 



192 ELECTRICAL AND MAGNETIC CALCULATIONS. 



ampere should be taken for armature conductors, and 
800 to 1000 circular mils per ampere for the fields. 

(3) The proper I 2 R losses for armatures and fields 
may be arrived at from the following table derived from 
several standard modern machines of various sizes. 

7. Table of I 2 J? Armature and Field Losses. 



Size. 


Armature 
Loss %. 


Field Loss %. 


100 K.W. 

50 K.W\ 

25 K.W. 

5 K.W. 


2.0 

2,2 

2-5 
4.0 


I.I 
1.6 

2.5 
4-5 



(4) Peripheral speeds of armatures vary from about 3000 
feet per minute as an average in ordinary sizes, to 5000 
feet per minute or more in very large sizes such as 
Ferranti disk armatures. 

( 5 ) The angle of span of the poles in two pole dynamos 
varies from 120 to 145°. An educated judgment as to the 
proper separation of the pole tips so as to reduce leakage 
can be relied on in this particular. Poles may cover 
about three-fourths of armature surface. Say 0.6 to 0.75. 

(6) The armature laminae vary in thickness from 25 to 
50 mils ; say 30 mils as an average. Usually 10 per cent 
to 20 per cent may be allowed as a reduction in the arma- 
ture cross section for insulation between laminae and 
shaft space. 

( 7 ) For armature proportions we may take the ratio of 
the external to the internal diameter 10 to 8 ; sometimes 
10 to 7 is used. For drum disks the ratio is about 10 to 
3. These values are not rigid. Different values would 



ELEMENTS OF DYNAMO DESIGN. 1 93 

perhaps be assigned by different designers. The ratios 

of drum armature lengths to diameters vary considerably. 

But representing the length by / and the diameter by d, 

II 
the ratios are most often— = 1, — = 2, or some ratio be- 

d a 

tween these. For ring armatures the ratios vary from 

/ / 

-d=* t0 -d = 2 - 

( 8 ) Flux densities in armatures for constant potential 
direct current machines vary between 10,000 gausses and 
15,000 gausses. However, it is not often advisable to go 
beyond 12,000 gausses. For arc machines the density 
often reaches 18,000 gausses. Perhaps 16,000 gausses 
will represent the average flux density for such machines. 
For alternators the density is considerably less, varying 
from 6000 to 7000 gausses, while in coreless disks it is 
about 5000 gausses. 

Air gap intensities vary from 3000 to 7000 gausses in 
direct current machines, and from 2500 to 5000 in alter- 
nators. 

Field densities vary from 12,000 to 17,000 in both direct 
constant potential machines and alternators. It will be 
about 18,000 gausses in arc machines. These values are 
all for wrought iron. When cast iron fields are used 
6000 to 8000 gausses is as high as permissible in any 
type. 

(g) For relative cross section of field and armature 
Thompson gives the following ratios of field to armature ; 
ring machines, wrought field, 1.66; cast field, 3. For 



194 ELECTRICAL AND MAGNETIC CALCULATIONS. 



drum armatures, wrought field, 1.25; cast field, 2.3. 
These, however, are only tentative, being subject to con 
siderable variation. 

55. Armature Magnetization. — In the field calculation 
of Chapter X. no account was taken of armature re- 
action due to the magnetizing effect of the armature cur- 
rent. In the design of dynamos allowance must be made 
for this armature magnetization, otherwise the voltage 
will not be high enough, nor constant at that ; for the 
larger the armature current the greater will be its effect 
in neutralizing and distorting the field. 



H K 




Fig. 19. 

In most machines there are two effects of the cur- 
rent in the armature conductors. Referring to Fig. 19, 
the direction of the field magnetization is represented by 
the darts on the poles, and the two effects of armature 
current by the darts on the armature, showing the direc- 
tion of the lines of force ; those on the right completing 
their circuit through one half of the armature and the 
iV-pole ; of course, there is a similar circuit on the left. 



ELEMENTS OF DYNAMO DESIGN. 1 95 

The brushes stand in the plane, CC called the commutating 
plane ; HOC is double the angle of lead, which is KOC ; 
LK is the normal plane, and if there were no skewing 
effect due to armature reaction, it would be the plane 
containing the brushes. The turns in the angle HOC, 
four in this case, set up an armature magnetization 
directly opposed to that of the field. These turns are 
therefore called back ttirns. The rest of the armature 
turns, approximately those coming under the span of the 
poles, set up a magnetization whose lines tend to cross 
those due to the field, and whose poles in the armature 
tend to be at right angles to those induced in the arma- 
ture by the field. These armature turns are called cross 
turns. Since there cannot exist two independent sets of 
poles in the armature core, there will be a resultant of 
the induced magnetization N X S X and the cross magnetism 
IPS', along the line JVS diagonally between the com- 
ponents. The combined effect is apparently a pulling 
down of the lines in one pole and a pulling up in the 
other in the direction . of revolution ; and the resultant 
magnetism passes diagonally up through the armature 
core instead of horizontally as it would were there no 
armature magnetism. 

The ampere-turns on the armature causing the back 
magnetization are 

Back ampere-turns = n h X \I ( IO °) 

in which n b is the number of conductors within the double 
angle of lead. In the figure n b = 4. Also / is the 
whole current output of the machine. 



196 ELECTRICAL AND MAGNETIC CALCULATIONS. 

The ampere-turns on the armature causing the cross 
magnetization are 

Cross ampere-turns == n c X \ /, ( IQI ) 

in which n c is the number of conductors on one-half the 
armature less n b , the number in the double angle of lead. 
In other words n c is the number counted within the 
angle COP or COH. I, as before, is the total armature 
current. We take \ I, since each conductor carries but 
one-half the total current. 

Suppose, as in the figure, that there are 14 cross turns 
and 20 amperes output. The back ampere-turns are 
4 X \ X 20 = 40, and the cross ampere-turns are 
14 X \ X 20 = 140. 

If no other means of knowing what portion of the sur- 
face turns are back and what cross, an approximation may 
be made by taking, the number between the pole tips as 
the back turns and those within the span of the pole as 
cross turns. 

Evidently when the angle of lead is known the number 
of turns within double this angle is readily found. For 
example, let a be the angle of lead and n the whole num- 
ber of armature turns ; then 

2 a a 

180 90 

Let a= 1 5 . 

Then n h = -2— = - ■ 

90 6 

56. Compounding. — The shunt field winding is the 
only one effective when the machine is on open circuit ; 
that is, when it is running at normal speed, but unloaded. 



ELEMENTS OF DYNAMO DESIGN K)J 

Therefore the regulation must be effected by compound- 
ing the dynamo ; that is to say, by adding coils on the 
cores in series with the circuit and with the armature. 
These series coils must compensate for the drop in the 
armature resistance and in their own resistance ; they 
must also compensate for the effects of the back and the 
cross magnetizations ; they must in addition compensate 
for the effect of the increased saturation of the fields and 
armature when the machine is loaded. Furthermore in 
the case of over compounding, still more series turns must 
be added to compensate for a certain amount of line 
loss. 

57. Useful Formula. — Volts drop : let R a = the 
armature resistance, / the armature current, E a = the 
volts drop in the armature. Then 

E a = RJ. (102) 

If R s is the field resistance, I s = the field current,' E s = 

the field drop, then 

E s = RJ S . (103) 

Since the shunt winding and the external circuit are in 
parallel, E s is the same as the external, or line E.M.F. 
The current in the field winding is the total current times 
the shunt field loss per cent. This varies somewhat with 
the size of the machine; see Sec. 54* But in ordinary 
sizes it will be about 2.5%. 

The series field drop will be 

E f =R f I e . (104) 

R f is the resistance of the series winding, I e is the current 



198 ELECTRICAL AND MAGNETIC CALCULATIONS. 

carried to the external circuit ; it is the total / less the 
shunt field exciting current I s . In other words, 

/=/« + /.. (105) 

It will make but little difference in the temperature of the 
armature to design its coils for the external current, mak- 
ing it carry in operation the extra amperes necessary for 
the shunt excitation. 

The E.M.F. to be generated in the armature coils is 

E = E a + E f +E e . (106) 

Strictly speaking, E a includes the drop due to cross and 
back turns as well as the IR drop in the armature con- 
ductors. 

The e?iergy lost in the armature is 

W a =£ a x7=(£ -E e - £/) I. (107) 

E e is the terminal or line E.M.F. 

The watts lost in the shunt field are found by 

W s = EJ S = EJ S = RJ S \ (108) 

The series field loss is given by 

W f =E f I t = R f I?. (109) 

The eddy current* loss in the armature, from Stein- 
metz's formula, is 

W e = V(tfbf X IP" 16 . (no) 

See Sec. 42 for (no). Several conditions may render 
the results of the actual tests on the dynamo widely 
diverse from these given in (no). Yet the latter will 
serve very well as a check. 

* See Equation (58). 



ELEMENTS OF DYNAMO DESIGN 



I 99 



Hysteresis losses may be approximated by 

W h = Vhf&\ (in) 

where the hysteretic constant h is known for the quality 
of iron used in the armature. Taking the following in- 
ductions in the armature, the hysteresis losses correspond- 
ing for soft iron or soft steel will be about as tabulated. 

8. Table of Hysteresis Losses. 



B. 


Loss in Watts. 


IO.OOO 
12,000 
14,000 
15,000 


Wk= Vfx 6.28 x io-4 
Wn— Vfx 8.40 x io~ 4 
Wh = Vfx 1.08 x io -3 

JVh — VfX 1.20 X I0~ 3 



V is the volume of iron in cubic centimeters and f is the 
number of cycles per second. 
The electrical efficiency is 

Elec. eff. = 5^4*- (112) 

The commercial efficiency, while it must be finally settled 
by experimental test, may be approximated by adding to 
the available output of the machine all the losses as just 
given, then divide the output by the sum. Hence 

Output 



Com. eff. = 



(»3> 



Output and losses 

It is seen that no account is here taken of windage, 
bearing and brush friction, all of which reduce the effi- 
ciency, and which can be determined by actual test. 

*58. Illustrative Examples. — 1. Let it be required to 
design a half horse-power shunt-wound motor for no volt 
circuits. 

* These examples should also be worked out from gilberts per centimeter (H) 
taken from curves on p. 229, or from ampere-turns per centimeter taken from tables 
on p. 298. 



200 ELECTRICAL AND MAGNETIC CALCULATIONS. 

As small a machine as this, especially if it runs light 
part of the time, cannot be expected to supply its power at 
greater than 60% to 70% efficiency. Hence the intake 

will be, say v- 0.60 = 625 watts. At no volts there 

will be 5.6 amperes of current. Allow 400 circular mils 

per ampere, requiring — x 400 =1120 circular mils. 

This is nearest No. 19. We may approximate 10% PR 
losses in fields and in the armature, making each 62.5 
watts lost. Suppose we use the allowable temperature 
rise as a basis for solving the armature in this case. If we 
allow 1 watt loss per square inch of external surface, we 
obtain the size as follows : let d = the core diameter, 
/ = its length. Consider that these shall be equal. The 
end surfaces are 2 d 2 X 0.7854. The convex surface is 
d X 7T X /. Since d = /, 

2 / 2 X 0.7854 + ?r/ 2 = 62.5, 

62. c 
and P = — — • Whence / = 4 inches. 

4.7 
For the fields the current will be 

10% of 5.6 = 0.56 ampere. 

This at 800 circular mils per ampere requires 448 cir- 
cular mils, or No. 24 wire, having 26 ohms per 1000 feet 
cold. Assuming a rise in temperature of 40 C above the 
air, and taking 1 % increase for each 2-J-° rise, the resis- 
tance, hot, will become about 116% of 26 = 30 ohms per 
1000 feet. The field coils must have a resistance of 
no -f- 0.56 = 196 ohms. Therefore the length = 196 

+ 3° = 6 533 feet - 



ELEMENTS OF DYNAMO DESIGN 201 

d>nv 
From (79) E = -^—g- • Since E = no, taking v = 30 

r.p.s., assuming ^ = 6000 gausses, and 70% available 
armature cross-section, we have 

no x io 8 

n = t rs 7 = 9°° wires - 

(4 x 2.S4) 2 X 0.70 X 3° X 6000 

Take, say 20 commutator segments, thus requiring 900 -r- 
20 = 45 wires per coil and per slot, and 22 turns per 
coil. Divide the circumference of the armature into 40 
parts, half of it being for slots. Circumference = 4?r = 
12.56 inches. Hence each slot will be 12.56 -r- 40 = 0.3 
inch wide. No. 19 wire is 35.4 mils bare, or about 50 
mils d.c.c. This will allow 0.31 -*- 0.05 = 6 wires in one 
layer in each slot ; and the slots must then be 44 h- 6 = 8 
layers deep. The wire space in depth = 0.05 X 8 = 0.4 
inch. The slots must be insulated on the sides and 
bottom, say to the thickness of one wire. Hence 5 wires 
wide will require 9 wires deep + space for insulation = 10 
wires deep. Depth therefore = 0.05 X 10 = 0.5 inch. 
This will perhaps be also sufficient to allow for bond-wire 
space. 

The total length of wire on the armature will be about 
750 feet, allowing for piling up on the ends and for com- 
mutator connections. No. 19 has 8.5 ohms per 1000 feet. 
Hence the I 2 E loss will be 5.6 s X 1.6 13 = 50 watts, or a 
little less than 10% as assumed at the outset. We get 
1.6 1 3 as the hot resistance of 750 feet -7- by 4 for the 
maximum armature resistance. 

We may take the laminae for the core 0.04 inch thick. 
The volume of the core = 10 x 0.7854 x 10 = 785.4 cc. 



202 ELECTRICAL AND MAGNETIC CALCULATIONS. 

Applying table 8 under (in), taking the constant at 5 
and calculating the hysteresis loss, we have 

W h = Vfx 5 x io- 4 = 785.4 X 30x5 X 10- 4 
= 12 watts. 

Also for eddy currents from (no) 

W e = V(tfl?) 2 X 10- 16 = 785.4(40 X 30 X 6000) 2 X io~ 16 
= 4 watts. 

The total armature loss is 50+12 + 4 = 66. 

But since only half the external surface is iron the flux 
density will be doubled in the teeth, so that both hysteresis 
and eddy currents will be greater in the teeth. If neces- 
sary the volume of the teeth may readily be determined, 
also the volume of the rest of the core, with their corre- 
sponding flux densities, and the iron losses for these parts 
determined separately. This would perhaps increase the 
above value to 70 watts, or more. 

Calculation of Field. — Assume the coefficient of leak- 
age to be v = 1.5, thus making the lines of force in the 
cores of the field magnets become 

<f> f = 70 x 6000 x 1.5 = 630,000 maxwells. 

We shall make the core of wrought iron, but for purposes 
of rigidity it will be better to make it larger than other- 
wise necessary. Hence take B = 8000 gausses. Whence 

D of cores =4/6^000 _ = IO cm . = 4 in . 

▼ 8000 

Now if the poles and core are all of wrought iron 4 inches 
will be ample ; but if cast iron be used the area of cross 
section must be somewhat larger so as to reduce B f to 



ELEMENTS OF DYNAMO DESIGN 203 

6000 gausses, perhaps. Use single yoke core. Allow a 
clearance of 0.08 inch, or 0.16 inch both sides. 

Now to approximate the length of the magnetic circuit: 
from middle of pole face through pole to a point on a 
level with under side of the armature, say 3 inches ; 
thence to top of field-winding, say 1 inch ; thence to 
center of core, that is, wire and half of core depth, say 4 
inches ; yoke, say 4 inches, core length, making total 
approximate field circuit 20 inches. Also make poles of 
such shape and dimensions as that their cross-section will 
be approximately equal to that of the armature = 4 X 
4=16 sq. inches =100 sq. cms. 

Now for the field core the reluctivity is 
a 0.0004 

1 — bB 1 —0.000057 x 8000 

= 0.00073 oersted per cc. 

_ _ / 10.16 . 

0t c = k - = 0.0007 3 x — = 0.000004 oersted. 

a ,0 78.54 ** 

The rest of the field will be 16 inches = 40.64 cms. long 
and 16 sq. inches = 103.25 sq. cms. section. 

tx r> 630,000 

Hence J3 f = = 6000 gausses. 

103.25 

And k = 0.0006 oersted per cc. 

r 40.64 
(R. f = 0.0006 x = 0.00023 oersted. 

io3- 2 S 
For the air gap, assuming all the lines to pass through it, 
the reluctance will be 

0.4 
(JL = = 0.004 oersted. 

y 100 



204 ELECTRICAL AND MAGNETIC CALCULATIONS. 

Hence, neglecting the reluctance of the armature, the 
total reluctance will be 

61, = 6L P + 61^ + 6l c = 0.004324 oersted. 

The approximate ampere-turns are 

<£6l 630,000 X .004324 
nl = — = - = 2170. 

0.47T 1-256 

Therefore the number of turns will be 

2 1 70 

n = — — = 3875 turns of No. 24 wire. 

This will have a diameter of 35 mils d.c.c. The length 
of wire already found is 6533 feet. Allow 20 per cent of 
this for the regulating rheostat, leaving 

80% of 6533 = 5226.4 feet for field wire. 

The mean length of one turn will be 5226.4-7-3875 = 
1.35 ft. mean circumference of winding. The mean 

j -2 r V j2 

diameter will be -^ = 5.2 inches, making the 

3- J 4 

depth of wire 5.2 — 4 = 1.2 inches. Allow 0.1 inch for 
core insulation, and we have for wire space, 1.2 — 0.1 = 
1.1 inches, requiring 1.1-7-0.035=31 layers deep. There- 
fore the cores must be long enough ^3875-7-31 = 125 
turns in one layer. The length for wire space is then 

125 X 0.035 = 4-38 inches. 

Allow I inch for fiber collars at each end of core ; this 
makes the total wire space 3.75 inches, shortage being 
made up in greater depth, if the four-inch core be used. 



ELEMENTS OF DYNAMO DESIGN. 20$ 

To correct our data. The core reluctance will not be 

appreciably changed, since the corrected core length is 

only a little different from that assumed. The diameter 

of pole bore is 4 + 0.16 = 4.16 inches, circumference 

= 13.06 inches. Choose the angle of span = 120 , thus 

, . , - r 1 r I2 ° 13.06 

making the length of curve of pole face = -r— X — — = 
00 180 2 

4.3 inches. The separation of pole tips will then be 

13.06 — 8.6 . , _ , , 

-^ = 2.23 inches. Pole face area = 4.3 X 4 = 

2 

17.2 sq. inches. We may so shape the poles that the 
average area will be about 16 sq. inches as assumed, and 
so no correction for field area need be taken. 

Now to obtain exact length of flux lines. From the 
center of pole face horizontally 2 inches ; vertically down 
3 inches to top of winding ; 1.6 inch depth of winding; 
2 inches to core center. The total = 6.6 inches to a 
horizontal line through the center of core. Thence say 
2.5 inches inward to core end, making 11.1 inches on one 
side exclusive of core. But the curving of the flux paths 
in the field parts will reduce this to say 10 inches, or 
20 inches on both sides. Our assumed length was 
16 inches. Hence the reluctance of this portion will be 
I of the assumed value. Therefore f of 0.00023 = 
0.00029 oersted. The total reluctance will now be 
0.004388, and the ampere-turns 2200 instead of 2170. 
Hence an additional layer of wire on the core will be 
sufficient, especially since our calculations assume that 
the total core flux passes through the complete field 
circuit. 

Now a little time on the drawing-board* will suffice to 



206 ELECTRICAL AND MAGNETIC CALCULATIONS. 

represent the machine to scale, with the proper forming 
of the outline so as to present a satisfactory appearance 
and to save any iron possible whose removal will not 
prejudice the operation of the machine. Show substan- 
tial arms cast on the poles to support the armature bear- 
ings, and design feet to bolt on slide base. 

Summary: — Armature 4" X 4" 5 speed 1800 r.p.m. 
Current 5.6 amps. ; R a = 1.61 ohms. 
Wire length 750 feet ; size No. 19 d.c.c. 
Coils 20 having 22 turns each. 
Laminae 100, 0.04 inch thick. 
Shaft |". 

Fields, core 4" long, 4" diameter. 
Area 13 sq. in.; wire space 3.75" in. 
Flux length 10 inches each limb. 
Total flux path 24". 
Current 0.56 ampere ; turns 3875. 
Wire length 5300 feet, No. 24 wire. 
Span of poles 120 ; separation 2.3' 
Losses, armature/ 2 ^ =50 watts. 
" Iron = 20 " 
" Total = 70 " 
Field 7 2 i?= 62 " 
Total =132 " 



jr 



Efficiency = ^ + ^ + 132) = 74%. 



2. Required to design a 5 K.W. bi-polar direct current 
machine for 120 volts full load. 

As a check suppose we determine the size of the 
armature in two ways. First taking the peripheral speed 
at say 3600 feet per minute ; also 1800 revolutions per 



ELEMENTS OF DYNAMO DESIGN. 20J 

minute as a good speed for this size of machine. The 

3600 2 
diameter of the armature must be -^ = - ft. = 8 inches. 

__ , . . . - n . . IO007T 7T 

Take also a length or 8 inches. 

Second, if we take i-J- sq. inches per watt of I 2 R loss 
which according to the table will be 4% for a 5 K.W. 
dynamo, the diameter of the armature will come out as 
follows : d 2 X 7T + 2d 2 X 0.7854 = 1.5 X 200. 
d = 8 inches, nearly. 

We shall therefore work on the basis of an armature 
8" X 8", running at 1800 r.p.m. = 3780 feet per minute, 
peripheral velocity. 

The armature current is 

cooo 

I a = = 41.7, say 42 amperes. 

120 

At 600 circular mils per ampere, the wire must be — 

X 600 = 12,600 cir. mils, which is between Nos. 10 and 9. 
Take No. 10 = 10,381 cir. mils. = 500 circular mils per 
ampere. 

If we choose 4 volts between adjacent commutator bars 
there will be 

2 x ( ) = 60 commutator bars, 



I2o\ 

~4/ 



This might be made 50. Also 

j£Xio 8 E X io 8 

n = = • 

<t>v BAv 

Deducting 20% from armature cross section, 

A = 64 X 0.80 = 51.2 sq. in. = 330.24 sq. cms. 

Hence ati? a = 10,000 gausses, <£ a = 3,302,400 maxwells, 

So that Q 

120 X io 8 

// = =120 conductors = 60 coils. 

3,302,400 x 30 



208 ELECTRICAL AND MAGNETIC CALCULATIONS. 

If we choose 60 commutator bars there will be 1 turn in 
each coil. If we make 60 slots on the armature the wires 
will be 2 deep in each. Let the slots occupy half the 

O y* 

surface; then each will be = 200 mils wide. Now 

120 * 

No. 10 wire = 10 1.4 mils uncovered, or about 120 mils 
d.c.c. The sides and bottom of the slots must be fibered, 
say 44.5 mils each to make up to about 209 mils; i.e., 
120 + 2 X 44.5 = 209. Two wires deep make 240 mils, 
which added to, say, 50 mils insulation in the bottom, 
make 290 mils. Further, the bond-wires, say No. 18 = 
40 mils, and the insulation under them, say 20 mils, make 
an additional depth of 60 mils, thus requiring the slots to 
be 350 mils deep. 

Now if we may assume that the flux lines enter the 
teeth only, their density in the teeth will thus become 
20,000 gausses, since only half the surface is iron. 

The I 2 R a loss may now be estimated as follows : mean 
turn = (8 — 0.35) X 2 + (8 X 2) = 31.3 inches. Allow 
25 per cent for rounding the shaft and piling up and 
making commutator connections, making the total length 
of wire on the armature equal 

31.3 X 1.2c; x 60 , r 

2— 2 5 a 195.625 feet. 

No. 10 has a resistance of 1.023 ohms per 1000 feet. 
Hence our armature wire will have a resistance of 
0.1956 X 1.023 = °- 2 oh m » The resistance of the arma- 
ture is therefore \ of 0.2 = 0.05 ohm cold. To allow 
40 C. rise will increase this about 16%, making 0.06 
ohm hot. Hence 

PR a = (42 + 4.5% of 42) 2 X 0,06 = 116 watts. 



ELEMENTS OF DYNAMO DESIGN 209 

We allow the armature current to increase 4.5 per cent to 
supply the field current. The per cent loss will now be 
116 -*- 5000 = 2.5%, or somewhat less than the tabular 
value for machines of similar size. 

To calculate the fields. Take two cores, wrought iron ; 
use cast iron poles, and the cast iron bed-plate for yoke. 
The leakage coefficient we shall assume to be v = 1.4 to 
be on the safe side. <f>/= 3,302,000 X 1.4 = 4,623,000 
maxwells. Allowing B c = 15,000 gausses, the core diameter 
will be 



_, . /4,62^,000-H K,000 n . , 

D c = I/ - — - = 19.8 cms. = 7.7 inches. 

V 0.7854 

Consider 4,000,000 lines to pass through the poles and 
B p = 8000 ; then the area of poles will be 

A p = 500 sq. cms. 

Since the length of the pole bore = 8 inches = 20.32 cms., 
the curve of pole face must be 500 -^ 20.32 = 24.5 cms. 
Make the clearance -| inch, thus requiring the circum- 
ference of polar bore to be 8.25 X -k = 25.9 inches = 
65.78 cms. Therefore the separation of pole tips = 

65.78 — 24.5 X 2 . 

-^-j. ^ = 8.39 cm s. = 2>-?> inches. The angle 

of span is , 9 X 180 = 134 . We shall let this stand 
65.78 

as satisfactory. However, it would be possible to in- 
crease polar density somewhat, thus reducing the angle 
of span, which in turn would tend to reduce leakage. 

There will, of course, be ample cross section in the bed- 
plate; hence B y will perhaps not exceed 5000 gausses, or 



2IO ELECTRICAL AND MAGNETIC CALCULATIONS. 

6000 at the most. At 2 amperes exciting current the 
field loss is 

I^Rf = 120 x 2 = 240 watts. 

There will be 120 watts dissipated in each coil which, if 
we use the rule of 2 square inches per watt, will have 
240 square inches external radiating surface. 

As a first approximate solution take § inch as depth 
of wire on field core. The length of the core will 
be 240 -r- [(7.7 + 2 X I) X 7r] = 8 inches = 20.32 cms. 
The average length of the pole will be about 5 inches = 
12.7 cms. The finished coils should stand about 4 inches 
apart; therefore the yoke length will be about 13 inches 

= 33 cms - 

For cores, 

_ 0.0004 /- 

k = = 0.00276 ; 

1 — 0.000057 X 15,000 

20 3 2 X 2 

(R c = 0.00276 X — - — 5 = 0.000364. 

308 

For poles, 

. 0.0026 . 

k = = 0.0102 oersted per cc; 

1 — 0.000093 x 8000 

1 2 1 X 2 

(R„ == 0.0102 x — = 0.000C2 oersted. 

500 * 

For yoke, 

0.0026 ^r,r,rZ%. 

k = = 0.00C00 ; 

1 — 0.000093 x 6000 

6i y = 0.00588 x 7^— = 0.00032 oersted. 

Next to find the armature reluctance, first find that of 
the teeth at B = 20,000, then of the rest of the armature 



ELEMENTS OF DYNAMO DESIGN 211 

body at B = 10,000. The number of teeth under the 
pole = \\% x 30 = 22 teeth whose area = 22 x 0.209 x 
2.54 X (8 X 2.54) X 0.80 = 190 sq. cms. Take ^ = 30 

from table 13 at B = 20,000; £ = -=0.033. ®-t—°'°33 

°' 88 9 X 2 oat r 

X = 0.000308. Area of rest of armature = 

190 

(20.32 — 1.778) X 20.32 X 0.80 = 301.38 sq. cm. Length 

of lines through this, say 20 cms., allowing for curving 

round the shaft. 

B a = 3,302,400-7-301.38 = 10,000, approximately. 

_ 0.0004 

k a = = 0.00093 ; 

1 — 0.000057 x 10 

(ft = 0.00003 x = 0.00006 oersted. 

301 

Diam. of polar bore = 65.78 -r- 3.14 = 20.95 cms. 
Length of air gap = 20.95 ~~ 2 °-3 2 = 0-63 cm. 

. . soo + 190 
Area air gap = — =345 square cms. 

Hence <JL = — — - = 0.00182 oersted. 

345 
Armature, 

3,302,400 (0.000308 -f- 0.00006) 

I.256 
= 9 68 ; n a = 4 8 4. 

Cores, 

4,623,000 X 0.000364 



(«/)c = 

Poles 



1.256 
= 1340 ; n c = 670. 



, „ 4,000,000 X O.OOOC2 ^ ^ n n 

( )p = ^^6 = l6s6 ; "* = 828 ' 



212 ELECTRICAL AND MAGNETIC CALCULATIONS. 

Yoke, 

. _ N 4,000,000 X 0.00032 

(«/% = — g =1018; ^=509. 

Air, 
/ r\ 3,302,400X0.00182 
(^). = — ^ = 47^5 ; n = 2393. 

Total ampere-turns = 9767 ; ?^ = 4883 turns. / 2 ^?/ = 
240 watts; R f = — g" = 6° ohms. The core diameter = 

7.7 inches ; assuming depth of wire \ inch, mean length 

tt(7.7 + 0.87c) 

of one turn = w ' ^ = 2.24 feet. 

12 

The total length of wire = 2.24 x 4883 = 10,938 feet. 
Hence R f per 1000 feet = 60 -r- 10.938 = 5.5 ohms, cor- 
responding approximately to No. 17 = 2048 circular mils, 
making 1024 circular mils per ampere of exciting current. 
No. 17 wire = 45.3 mils bare =55 mils d.c.c. Allowing \ 
inch for collars on the core, there will be in one layer 

o _ n r 

^. = 136. Hence 2442 turns on each core require 18 

layers deep = 0.99 inch. This exceeds \ inch given for 
the depth of wire ; but the core need not be lengthened, 
since only 18 layers deep are required, (0.99-^0.055). In 
one layer there will then be 2442 -*- 18 = 136 turns = 
136 X 0.055 = 7-5 inches for winding space. Adding 
\ inch at each end for a collar, we have 8 inches for total 
core length. Now this will not change the core reluctance 
nor ampere-turns from their assumed values of 0.000364 
and 670 respectively. The total turns are 4896, or 2448 on 
each core. This again will make a slight difference in the 
resistance, which will now be 4896 X 2.24= 10.97 thousand 
feet having 5.18 X 10.97 =56.8 ohms; assuming 40 C. 



ELEMENTS OF DYNAMO DESIGN 21 3 

rise, thus increasing the resistance by 16%, we have 
65.8 ohms for the resistance of the shunt fields. This, 
however, can be reduced 10% to 15% to allow for rheo- 
stat resistance, bringing field resistance to about 60 ohms. 
This calculation will not give constant potential under 
changing loads. Hence the series compensating turns 
must be determined. These must compensate the cross 
turns, and back turns, the armature drop, and the series 
field drop. Now by reference to Sec. 55 it will be seen 
that the cross ampere-turns on the armature produce a 
magnetism at right angles to that of the field. It will be 
nearly correct to consider that the resultant of the shunt 
field ampere-turns and the armature cross turns will be 
the whole number required. In other words, the differ- 
ence between this resultant and the shunt ampere-turns 
will nearly represent the additional series ampere-turns to 
compensate for cross magnetization. 

Cross turns = — ^— X 60 = 4c;. 

' 44 

Cross amp.-turns \ I X n e = — X 45 = 990. 

The armature current is here taken as 44 amperes to 
include the shunt field exciting current, 2 amperes. 

Back amp.-turns (60 — 45) X 22 = 330. 

f ' 2 1 

Cross. mag. r comp. amp.-turns = (9792" + ggo")^ 

= 9792 + 50. 

Series turns = — = 1 i turns for cross effect. 
42 F 

Series turns = ^— = 7f turns for back effect. 
42 

Armature drop = 44 X 0.06 = 2.64 volts = 2.2 c / 
of normal E.M.F. 



214 ELECTRICAL AND MAGNETIC CALCULATIONS. 

Hence 2.2% of shunt ampere-turns are required to com- 
pensate it; 2.2% of 9792 = 216 ampere-turns, requiring 
series turns = 216-7-42 = Sj turns for armature drop. 

Total series turns = 1^ + 71+51=15. The wire should 
be taken large enough to reduce the drop in the series 
field coils to a minimum. Assume 1000 circular mils per 
ampere. The wire must then have 42 x 1000 = 42,000 
cir. mils, between No. 4 and No. 3. Say we choose 
No. 4 = 0.2 inch bare = about 0.25 inch d.c.c. The 
winding space then is 15 X 0.25 = 4 inches, or 2 inches 

t. t 4.1, * • (7-7 + 1.98) ir X 15 
on each core. Length of wire = ^^— - ^ = 

38 feet having a resistance of 0.0095 ohm. 

Series field drop = 0.0095 X 42 = 0.40 volt. 

This is so small that the excess turns taken for compound- 
ing effects, and the margin on the shunt field will amply 
compensate. However, an extra series turn may be 
added, then the rheostat set to produce proper running 
voltage. 

Now if it is desired to overcompound the machine, 
say 5 °f , an additional number of series ampere-turns may 
be put on equal to 5 % of the shunt field ampere-turns. 
This machine would require 5% of 9792 = 490 ampere- 
turns, and say 12 series turns. 

If in designing, the induction allowed in the fields is 
kept well .within reasonable limits, the added series 
ampere-turns will not affect the saturation of the iron 
sufficiently to lower appreciably its permeability. 

Summary : Machine 120 volts, 42 amperes. 

Armature, 8" X 8"; 1800 r.p.m. ; B a = 10,000. 



ELEMENTS OF DYNAMO DESIGN 21 5 

I a — 44; wire No. 10, 60 coils, 60 slots. 
Wire length 195.6 feet; R a = 0.06. 
Volts drop 2.6 ; <j> a = 3.3024 X io 6 . 

Fields: Core diam. 7.7" ; B c == 15,000. 

<£ c = 4.623 X io 6 ; shunt turns 2448 each. 

Exciting current 2 amperes. 

Wire length 10,970 ft. ; size No. 17. 

R f = 60 ohms. 

Series turns 15 ; No. 4 wire ; length 38 ft. 

The efficiency, so far as we are able to calculate it, is 
found as follows : — 

Hysteresis, 

W h = 20.32 2 X 0.7854 X 20.32 X 6.28 X io -4 X 30 
= 125 watts. 
Eddy current, 

W e = 6589 (40 X 30 X 10,000)* X io -16 
= 95 watts. 
I 2 jR a =116 watts. 
Total armature loss = 336 watts. 
Series field loss = 0.0095 X 42 = 16.76 watts. 
Shunt field loss = 240 watts. 
Total field loss = 256.76 watts. 
Total losses = 592.76 watts. 
Per cent loss = 592.76 -4- (5000'+ 596.76) = 10%. 
Efficiency = 100— 10 = 90%. 

3. Determine the essential data for a 10-K.W., 250- 
volt, 4-pole, cylindrical ring armature, direct current 
machine. Series connect the armature conductors so 
that only two brushes are used. 



2l6 ELECTRICAL AND MAGNETIC CALCULATIONS. 

By the methods already used, making proper allow- 
ances for internal ventilation, taking the I 2 R a loss equal 
to 3%, we obtain for the armature dimensions, length 
10 inches, outside diameter 10 inches, inside diameter 6 
inches, thus making the thickness of the ring 2 inches. 

Take a moderate speed, say 1200 r.p.m. 

$nvp 
Whence 250 = 



10" 

v8 



r , • , 2 50 X IO C 

from which d>n = ~ = 6.2 q X io 8 . 

^ 20 X 2 ° 

We may reasonably allow n = 200, thus giving 

<£ = 3,125,000 maxwells. 

The circumference = i07r= 31.416 inches. 

Take one-half this = 15.708 inches for slots. The 
machine current is 40 amperes, 20 amperes in each con- 
ductor, which at 500 circular mils per ampere requires the 
wire to be 

20 X 500 = 10,000 cir. mils = No. 10. 

This has an area of 7854 square mils. Take flat wire 
whose width = 3 times its thickness. For 7854 square 
mils the wire must therefore be 51 mils X 153 mils bare, 
and double cotton covered = 75 X 200 approximately. 
The slots must therefore be 

Width = 200 + 40 = 240 mils. 
Hence No. slots = 15.708 -5- 0.240 = 65, say 66. 

Putting 3 wires in each slot makes 198 surface conduc- 
tors. The speed must be slightly increased to compen- 



ELEMENTS OF DYNAMO DESIGN 21J 

sate for the two conductors dropped ; otherwise carry 
the field magnetization slightly higher. 

Depth of wire = 3 X 75 = 225 mils; allow 100 mils 
for insulation in the bottom of the slots, thus requiring 
the depth to be 325 mils. 

Exact width of each slot will be 

31.416 -h (66 X 2) = 0.238 inch. 

This is also the width of each tooth. 
The approximate length of wire will be 

66[(io + io + 2 + a )+6] = i6Sj say i?5 ^ 

No. 10 has 1.023 ohms per 1000 feet; hence 

_ (1.023 X 0.175) + 16% of (1.023 X 0.175) 

4 
= 0.0525 ohm. 

Allowing 0.1 inch clearance, the pole circle will have 
a diameter 

D = 10 + 0.2 =10.2 inches, and 

Circum. = io.27r = 32 inches. 

Making the width of the poles equal to their distance 
apart, we have 

Pole width 32 -r- 8 = 4 inches. 

Pole face length =10 inches. 
Area pole face = 10 x 4 = 40 sq. in. = 250 sq. cms. 

Pole flux, perhaps, 3.125 x io 6 X 1.25 = 4 X io 6 
maxwells. 

B p = 4 x io 6 -r25o= 16,000 gausses. 



2l8 ELECTRICAL AND MAGNETIC CALCULATIONS. 

This intensity may be reduced by making the width of 
the pole face slightly greater. To determine approxi- 
mately the field winding. Take the length of poles = 
6 inches = 15.24 cms. Also 

rr^i • 1 r f (4 X IO 6 ) "f" 12,000 

Thickness of pole ring = — '- = 7 cms. 

v & 10X2.54 ' 

= 2.5 inches. 

Mean circum. of ring (5.1 + 6 + 1^) X 2ir X 2.54 = 197 cms. 

Length of flux travel in pole ring is 

/= £ of 197 = 49 cms. 

For curving down to enter poles add 3 + 3=6 cms. 
making 55 centimeters. 

Armature, Norway iron, /= 10" = 25.4 cms., for lines. 

Area, A = (10 X 2.54) X (2 X 2.54) = 129 sq. cms. 

_ 3.125 X io 6 
B a = '- 129 = 12,100 gausses. 

O.OOOI , 

k = = 0.00035 oersted per cc. 

1 — 0.000059 X 12,100 

2 C.4. 
(R_ = 0.00035 x -^— = 0.00006 oersted. 
129 

Air gaps, length, / = 0.2 inch = 0.508 cm. 

A A 2 5° 

Area, A = -^— = 125 sq. cms. 

^ 0.C08 

(R„ = — - — = 0.004 oersted. 
125 

Cores, wrought iron, /= 15.25 cms. 
A = -^- = 125 sq. cms. 



ELEMENTS OF DYNAMO DESIGN. 2K) 

4 X io 6 

B r = = 16,000 gausses. 

c 2 X 125 & 

0.0004 i 

k = — = 0.004 c oersted per cc. 

1 — 0.000057 X 16,000 

j c 2 c X 2 

(ft = 0.0041: X = 0.0011 oersted. 

3 125 

Yoke 3 /= 55 cms. 

A = 175 sq. cms. 

4 X io 4 
B y = 5- 175 = 11,400 gausses. 

_ 0.0004 

# = = 0.00 1 14 oersted per cc. 

1—0.000057X11,400 

(ft y = 0.00144 X = 0.00045 oersted. 

Reluctance of a single circuit is 

(ft = 0.00006 + 0.004 + 0.0011 + 0.00045 
= 0.0056 oersted. 

Of two circuits in parallel the reluctance is 
(ft = 0.0028 oersted. 
M.M.F. = 0.0028 x 3.125 x io 6 = 8750 gilberts. 

Ampere-turns, each core, = — ^ — "- — = 3483. 

Exciting current I s = 3.5% of 40 = 1.5 amperes. 
Hence 

n = 2322, say 2325 turns, each core. 

Total turns, n = 2325 x 4 = 9300. 
2 CO 

R f = — — = 166.6 ohms = 41.6 ohms each core. 
1.5 



220 ELECTRICAL AND MAGNETIC CALCULATIONS. 

Assume for calculation that the depth of winding is i 
inch. Hence the mean turn = 32" = 2§ feet. 

Total length of wire each core = 6200 feet. 

The 41.6 ohms must be the hot resistance of the field, 
making about 37 ohms cold, which corresponds to No. 18 
wire = 40 mils bare = 55 mils d.c.c. 

Assume core length = 6 inches exclusive of collars and 
insulation. This will permit in one layer 

6 -5- 0.055 = 109 turns, and 

Number of layers deep = 2325-7-109 

= 21, and 36 turns over. 

Depth of winding = 0.055 X 22 = 1.2 inches. 



-48- 






A_J 



A 




Fig. 20. 

Hence it will not be necessary to modify our assumed 
dimensions except for symmetry in our drawing. Figs. 
20 and 21 show the general dimensions of the finished 
machine. 

Volts drop in the armature = 0.0525 X 40 = 2.1. 



ELEMENTS OF DYNAMO DESIGN 



221 



Hence the series ampere-turns to compensate will be 



2.1 . 

X 2325 = 20 on each core. 

250 

Number turns = — = ^ to each core. 
40 2 




Fig. ax. 

» 

To compensate cross and back turns will require an 
additional 5 % excitation. 

5% of 2325 = 117 ampere-turns. 



Hence 



IX 7 

n = — - = 3 turns each core. 
40 6 



This makes a total of about 4 series turns on each core. 

Length approximately = 3 X 16 = 50 feet, sav of No. 
2, thus giving 800 cir. mils per ampere. 

R = 0.050 X 0.16 = 0.008 ohm cold. 

Drop = 0.008 x 40 = 0.32 volt. 

The excess of series turns already put on will compensate 
for this drop. 

Calculate the efficiency of this machine and summarize 
the results of the problem. Calculate the overcom- 
pounding for 5 % line loss. 



222 ELECTRICAL AND MAGNETIC CALCULATIONS. 

4. Determine the data for a \ H.P. bipolar motor for 
112 volts direct current. 

In this design we shall express our dimensions in inches, 
and make use of table 13 and the curves in table 9 in 
order to illustrate their application. 

We have seen that reluctance 

(R = ^- = — • Also, (114) 

a a/jL 

nl= = 7 = - • (115) 

1.256 1.256 a/xX 1.256 x oy 

Here / is expressed in centimeters and a in square centi- 
meters. Now, if these dimensions are expressed in inches 
and square inches, the numerator of the fraction must be 
multiplied by 2.54, and the denominator multiplied by 
6.45. Introducing these values in the equation it becomes 

nf= <ft X / X 2 -54 = Q-3I3 2 X / X <fr 

" 1.256 X a X /x X 6.45 " a/ji ^ ' 

This gives the ampere-turns for any one portion of the 
magnetic circuit. So that knowing / and a we have only 
to determine fi from the flux density by reference to tables 
13 and 9. 

For \ H.P. at 112 volts the current required will be 
about 2 amperes, or one ampere in each armature con- 
ductor. For so small a machine we shall allow ample 
cross section in the armature conductors and select No. 
21, having 810 circular mils, diameter 0.034 in„, double 
cotton covered. Calculations as before, also reasonable 
assumptions for an armature of this power will give a 
diameter of about 3 inches. Take a length of 3 inches. 
Then for a speed of, say 2400 r.p.m., and say 16 slots 



ELEMENTS OE DYNAMO DESIGN 22$ 

holding 6 wires in width and 9 deep, or 54 wires each, 
and 54 X 16 = 864 conductors total, the required armature 
flux will be ■ io 8 X 112 

^ = 4ox864 =324 '° 74maXWells - 

This will make B about 60,000 per square inch. Making 
proper allowances for the insulation, the slots must be f 
inch wide by \ inch deep. Therefore the cross sectional 
area of the body of armature iron is 

3 * [3 — (I + f)] = 4.875 square inches, 

since the shaft will be f inch and the depth of two teeth 
is f inch. This will make the density 

324,074 -f- 4.875 = 66,476. 

The length of the flux lines in the armature body will be 
about 1.75 inches. 

The width of teeth at the top = 0.3125 inch; at the 
bottom, 0.1875 inch; average, 0.25 inch. 

Area = 0.25 X 3 X 6 = 4.5 sq. in. 

The density in the teeth = 324,074 -*- 4.5 = 72,016 lines 
per square inch. The average width of the tuft of lines 
entering each tooth, experience teaches, will be the width 
of the top of tooth plus the length of the air gap. Hence 
with 6 teeth under the pole at a time the air gap area 

will be (o. 3I2 5 + 0.062) x 3 X 6 = 6.741 sq. in. 

The field flux will be, say 324,074 x 1.1 = 356,481. 
We shall make the field rectangular in shape with in- 
wardly projecting horizontal poles. Make the poles 
square in cross section with slightly rounded corners, 
say 9 square inches area. The pole density will be 



224 ELECTRICAL AND MAGNETIC CALCULATIONS. 

356,481 -r- 9 = 39,609. Allow the wire to wind deeper in 
this small machine and shorten the poles so that about i-J- 
inches will be its depth. This will require about 1^- inches 
length of winding space. Adding \ inch each for collars 
makes if inches for length of poles, say ij 7 ^ inches to 
allow for portion outside of coil. 

From the poles the lines of force will have two parallel 
paths, and so the cross section of the yoke frame need 
be only one-half that of the poles. It will be 3 inches 
wide, of course, requiring a thickness of yoke of i-J- inches. 
We shall now have between ends of field coils 3 + tV + tV 
+ \ = 3^ inches. From center of yoke on one side 
through field coil, air gaps, field coil to center of yoke on 
the opposite side i|- + i| + 3i + i| = 8| inches. The 
height from center of horizontal portion of yoke (base) 
at the bottom to center of horizontal yoke at the top 
about 7f inches. The over-all length of the rectangle 
through the poles and armature will be 8^ + i-J- = 9I 
inches. Over all height = j% + 1% = g£ inches. 

We may now summarize and calculate the ampere-turns 
required for cast iron fields and wrought iron armature. 

Armature : I =1.75 inches; A = 4.875 square inches. 

fx, from table 13 for density = 66,475 

= 10,300 per sq. cm., is about 2000; 

1 r B 10,300 

also from o, u = -== = — - — 

y B 5 
= 2000. 

0.3132 X / X <f> 



a a ft 

= 18. 



a X fji 
0.3132 X 1.75 X 324,074 



4.875 X 2000 



ELEMENTS OF DYNAMO DESIGN 225 

Also using the column giving ampere-turns per cm. 

11 T 
length at B = 10,300, we obtain — - = 4, approximately. 

Hence (nf) a = (1.75 X 2.54) X 4 = 17.6, or 18 approxi- 
mately. 

Teeth : I = § inch ; A (under pole) = 4.5 sq. in. 

r 72,Ol6 

fx, for a density of — =11,200 gausses, is 1752. 

= 0.3132 x I x 324,074 = _ 
V ,a 4.5 X 1752 

Also for B = 11,200, — = 4.9, and (nf) t = 4.9 X (f X 

2.54) = 4.6. 

Air gap : 

I = y 1 ^ inch ; ^4 = 6.741 sq. in. ; /x = 1. 

/ n 0.3132 X 324,074 X 1 

(nI) Q = = 942. 

v Jg 6.741 X 1 X 16 ^ 

The density in air gap = 7454 gausses = H gilberts 
per cm. length. This is (650 X 10) + (20 X 44), approxi- 
mately, making 

— = (520 x 10) + (35 X 2 °) = 59°°> 

from table 13. 

nI 9 = 59°° X (A X 2.54) = 944. 

Poles : I = 2 inches average ; A = 9 sq. in. 

r i • 3Q>6oo 
/a, for a density = ^ = 6140, is 250, ap- 

proximately. 
Also from cast iron curve in 9 at B = 6140, H = 24, 

and B 6140 , .. 

u, = -== = = 2 co as before. 

r ZT 24 ° 



226 ELECTRICAL AND MAGNETIC CALCULATIONS. 

/ rN °-3 I 3 2 X 2 X 3^6,481 . . 

(nl) p = = 97-5> approximately. 

Also from table 13, (nl) p = 18.5 X (2 X 2.54) = 96. 

P^; /=8.25 inches from base of pole half way 
around each way ; A = 9 sq. in. for two 
parts in parallel. 

fiz= 250, as before, for density of 6140 gausses. 

= 0.3132 x 8.25 x 356,481 = 

. 9 X 250 

Also from 13, (nl) y = 18.5 X (8.25 X 2.54) = 388. 
Total ampere-turns in one-half of magnetic circuit = 
9 + 4.8 + 942 + 97.5 +402 = i455-3> sa Y ^S 6 - 

For both halves there are required 1456 X 2 = 2912. 

It will be noticed by use of the column giving the 
ampere-turns per cm. length, and multiplying by length of 
part, we get in general a different value from that just 
obtained. This is partly due to the approximation which 

Til 

has to be made to the value of — -> where it falls between 

the tabular values, and to differences in the iron forming 

the bases of the values in 13 and 9. 

Each pole will be wound for 1456 ampere-turns. For 

a depth of \\ inches the mean length of 1 turn = 1.169 

ft. We may then use for the size of the wire the 

formula, 

2 __ k X amp.-turns X mean length of 1 turn 

" E.M.F. ^ II7 ^ 

12 X 2012 X 1. 160 . „ 

= = ^co cir. mils. 

112 ^ 



ELEMENTS OF DYNAMO DESIGN 227 

k is taken at 12 instead of 10.79, to a U° w f° r a tem- 
perature rise of about 35 to 40 C. 

This size is between No. 25 and 24. Use No. 24. In 
one pound of No. 24 there are 724.64 feet of double 
cotton covered wire. Hence one pound will wind 

724.64 

— = 620 turns. 

1. 169 

No. 24 wire has a resistance of about 29 ohms per 
1000 feet, giving for 1 pound 

724.64 

x 29 = 21 ohms. 

1000 

For 1 pound at 1 1 2 volts, 

I= — = 5.33 amperes. 

Ampere-turns for 1 pound =620 x 5.33 = 3306, which 
is the same number that would be obtained with any 
number of pounds of the same size of wire, neglecting 
heating effect ; for as the weight of wire would increase 
the resistance would increase and the current reduce pro- 
portionately ; but the number of turns would be increased 
in the same proportion ; hence the ampere-turns would 
remain the same. 

To reduce current to a permissible amount for this size, 
as well as for economy, we may take 10 pounds, requir- 
ing 5*33 ~^ IO = °-533 amperes, and giving 620 X 10 = 
6200 turns, and 

6200 X 0.533 = 3306 ampere-turns. 

This is slightly larger than calculated, but it will give some 
margin for regulation by rheostat. Furthermore the rise 



228 ELECTRICAL AND MAGNETIC CALCULATIONS. 

of temperature will somewhat reduce the field current. 
Altogether this will make a very satisfactory winding. 

Calculate the depth of winding for the above number 
of turns. It will be approximately i| inches. 

Calculate the length and weight of wire for winding the 
armature, as previously indicated. 

Th,e pedestals for support of shaft will be f inch 
wide next to shaft, and 2 inches at the foot, with a pro- 
jection for bolting to base. They will be | inch thick 
and cut out to a web about \ inch thick. The arms on 
the base for support of pedestals will be, for the longer 
one on the commutator end, 3! inches long to center of 
pedestal, by 2^ inches wide; for the shorter one i|- inch 
to the center of pedestal, by 2^ inches wide. They will 
be cut out underneath, leaving only J inch of thickness of 
iron all around, and lag-bolted against the lower portion 
of yoke. The pedestals will be, from center of shaft to 
base, 3^ inches. The shaft is f inch through the arma- 
ture core and turned to | inch outside. It will be 9 
inches long. 

Length over all = 9I inches. 
Width over all = 9^ inches. 
Height over all = 9^ inches. 

Make a complete drawing of a motor of this design, in- 
cluding two elevations and plan of base and showing all 
necessary dimensions for its construction. Summarize 
all the work, giving all dimensions, sizes, weights, etc. 
Where necessary design exact details. For example, 
draw the details for the brush-holder, rocker arm, sixteen 
part commutator and pedestal. 



ELEMENTS OF DYNAMO DESIGN 2 29 



9. B-H CURVES 



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16000 
14000 
12000 
10000 

B 

8000 
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23O ELECTRICAL AND MAGNETIC CALCULATIONS. 



XII 



ALTERNATING CURRENTS. 

« 

59. General Definitions. — The nature of alternating 
currents necessitates the use of certain quantities which 
do not occur at all in direct current phenomena. Hence 
it will be well at the outset to gain some familiarity with 
these terms. 

(a) If the magnetic field in which a closed coil of 
wire revolves be uniform, the E.M.F. generated in the 
coil will vary as the sine of the angle through which it is 
turned, the rate being uniform ; in other words the curve 
of variation will be a sine curve ; the heights of the curve, 
or its ordinates, representing the E.M.F's. at the succes- 

B 




Fig. 22. 



sive points will be proportional to the sines of the corre- 
sponding angles through which the coil has turned from 
the position of zero E.M.F. The E.M.F. is zero when 



ALTERNATING CURRENTS. 23 I 

the plane of the loop is perpendicular to the lines of 
magnetic force. Fig. 22 shows the sine curve. Dis- 
tances measured horizontally represent angles, distances 
vertically represent E.M.F's. 

The height of the curve A r above the horizontal line 
represents the E.M.F. in the coil when revolved 50 from 
the position when the plane of the coil is normal to the 
lines of force. The height at B is proportional to the 
E.M.F. at 90 , or in the position when the coil's plane is 
parallel with the lines of magnetic force. This is clearly 
the position of maximum positive E.M.F. Therefore 

KM.F. at A' sin 50 

E.M.F. at B " = sin 90 " 

Hence the name sine curve. 

(p) It is clear from the curve that the average E.M.F. 
during one-half a revolution, or one alternation, is pro- 
portional to the average ordinate of the curve. Call the 
maximum ordinate at 90 one, then the area of the curve 
between the horizontal line and ABC is 2. The length 
AC = 180 = 7r = 3.1416. Therefore the average ordi- 
nate is 

area 2 /: c ^ 

i r = - = 0.637 of the maximum. 

length 7T 0/ 

Therefore, calling the maximum E.M.F. E, and the 
average E.M.F. E a , we have 

E a =o.6 37 E. (118) 

Suppose the maximum pressure be 100 volts, then the 
average pressure will be. 

E a — 0.637 x IO ° =63.7 volts. 



232 ELECTRICAL AND MAGNETIC CALCULATIONS. 

(c) The effective pressure of alternators is some greater 
than the average, and is that which is available for pro- 
ducing heat if applied to a resistance. The effective 
pressure is the square root of the mean of the squares of 
the successive pressures during one-half a revolution of 
the pressure coil, because the successive heats are pro- 
portional to the successive i? 2 's. The E.M.F's. vary from 
zero to i as a maximum. Therefore the mean of the 
squares is, by adding the squares of say 12 ordinates, 
and dividing by 12, J. 

Its square root is J I _ JL = 0.707 of the maximum. 

T 2 V2 

Calling the effective E.M.F. E e , we have 

Ee = O.707 E. (119) 

For a maximum pressure of 100 volts the effective E.M.F. 
will be (that indicated by a voltmeter) 

E e = 0.707 X 100 = 70.7 volts. 

(d) The period of an alternating current is the time 
from A in Fig. 22 to E, or the complete cycle, or one 
complete revolution of the coil in a 2 -pole machine. 
The number of complete periods per second is the fre- 
quency of the current. A cycle consists of two alterna- 
tions, or reversals of current. Therefore a frequency of 
125 per second means 125 X 2 = 250 alternations per 
second. The frequency of practical machines in the 
U. S. will nearly always be 60 or 25 per second, although 
125, 40, 66, 100 and 133 are used. A 4-pole machine 
gives a frequency equal to twice the number of revo- 
lutions per second; an 8-pole, 4 times the number. 



ALTERNATING CURRENTS. 



233 



Hence to obtain the frequency f of any machine, mul- 
tiply the number of revolutions per second by the number 
of pairs of poles p. An 8-pole machine running at 2400 
revolutions per minute has a frequency 

/= 2ioo X4=l6o . 

(e) An electromotive force of self-induction is set up in a 
circuit counter to the changing current caused to flow in 
the circuit. This counter E.M.F. is the result of the 
varying lines of force, set up by the variable current, 
which thread the adjacent portions or coils of the circuit. 
If this E.M.F. is set up by causing the lines of force to 
pass through the loops of an adjacent coil, the electro- 
motive force is that of mutual induction. The direction 




Fig. 23. 

of the induced E.M.F., whether of self-induction or of 
mutual induction, is, according to Lenz's law, in the 
direction to oppose the inducing force. This must also 
follow from the law of the conservation of energy; 
otherwise any empty circuit, say the secondary of a trans- 
former, would absorb as much energy as a loaded one. 



234 ELECTRICAL AND MAGNETIC CALCULATIONS. 

Since E.M.F. is proportional to the rate of change of the 
number of lines of force threading any circuit or loop, the 
self-induced pressure is the greatest when the alternating 
current is the least, or zero, and vice versa. Although at 
its maximum value the current sets up the maximum 
number of lines, the rate of change of the lines is the 
least ; while at zero current and lines the rate of change 
is a maximum, hence the induced E.M.F. is a maximum. 
A reference to Fig. 23 will make clear the relative 
values of the inducing current and E.M.F. of self-induc- 
tion at successive instants. It will be observed that the 
E.M.F. of self-induction E s lags 90 behind the curve of 
current and active E.M.F., E a . For an instant at B and 
D there is no change in the current, hence no induced 
E.M.F., while the change is most rapid through C and E 
shown by the slope of the curve, and hence there is a 
maximum counter E.M.F. at these points. 

CO Whenever an alternating pressure is applied to a 
circuit a certain component of it is useful to send the cur- 
rent through the resistance. This component is called 
the active pressure. The other component at right angles 
to the first goes to balance the E.M.F. of self-induction 
set up by the alternating current; this is sometimes 
called simply the self-induction pressure. The total ap- 
plied E.M.F is called simply the impressed pressure. This 
is analogous to a vessel moving, say N. of E. at a given 
impressed speed. It will be going east at a certain speed 
which we may call the active speed. We may suppose, 
in fact, the desired course is E., but on account of adverse 
currents or winds, the counter pressure, it has to steer N. 
of E. It will at the same time be moving north at a cer- 



ALTERNATING CURRENTS. 



235 



tain speed, which we may call briefly the counter or self- 
induction speed. These two are at right angles, or 90 
apart in phase. 

As already shown, the counter E.M.F. lags 90 behind the 
current ; therefore it is 90 back of the active pressure, that 
is, the phase difference is 90 , as shown in Fig. 23. 

The E.M.F. of self-induction being proportional to the 
rate of change of the current and active pressure is pro- 
portional to the angular velocity or displacement of the coil 
in which the induction takes place, or to 2 7r/,/ being the 
frequency ; it is also proportional to the coefficient of self- 
induction, a quantity depending upon the position, winding, 
etc., of the coil itself. Hence we can express the counter 
E.M.F. by writing 

E s =2irfLL (120) 

The resistance due to self-induction is 



R $ = 27T/Z. 

The active pressure is written 

E a = IR. 
in which R is the ohmic resistance of the circuit. 



(121) 
(122) 




Fig. 24. 

The relation of the different quantities in an inductive 
circuit may be expressed by a diagram, Fig. 24. The 



236 ELECTRICAL AND MAGNETIC CALCULATIONS. 

counter E.M.F. is represented by the vertical CB, and the 
active by AC, the impressed E { being the hypotenuse of 
the triangle of which the two components E a and E s are 
the two perpendicular sides. 

The angle BAC is the angle of lag of the current and 
active pressure behind the impressed pressure AB, 
which is 

^,= /V^ 2 + 4 7ry 2 Z 2 . (123) 

This is Ohm's law applied to alternating currents. 
Transposing this so as to put it in the usual form, we have 
for current in alternating current circuits, 



If L = o, then 1= — as in direct current circuits. 2 irf 

K 

is often represented by a>, the angular velocity ; in which 

case (124) would be written 



-w 



i + ^jr- ( I2 5) 

(g) If the values of the various E.M.F's. in Fig. 24 be 
divided by I, the current, the quotients will be the equiva- 
lent resistances as shown inside the diagram. R is the 
metallic or ohmic resistance ; 2 irfL is the inductive resis- 
tance, or reactance ; V47r 2 % /* 2 Z 2 + i? 2 is the apparent re- 
sistance or impedance. The hypotenuse, of course, is the 
square root of the sum of the squares of the two sides. 

( h ) In the above equations L is the coefficient of self- 
induction, a constant for any given circuit without iron. It 
is defined as the E.M.F. induced in a circuit wholly free 



ALTERNATING CURRENTS. 237 

from iron or other magnetic material, when the curre?it varies 
at the rate of one ampere per second. It is also called the 
inductance of the circuit. The unit of inductance is the 
henry, or that of 1 volt of counter E.M.F. when the current 
changes at the rate of 1 ampere per second. The henry 
= io 9 C.G.S. units of inductance. 

(*) The capacity of an alternating current circuit is 
the measure of the amount of electricity held by it when 
its terminals are at unit difference of potential. Every 
such circuit acts as a condenser, and a current will flow 
back and forth, even though it is open and in the ordinary 
sense unloaded, proportional to the rate of change of the 
active pressure of the circuit, and to the capacity of the 
circuit. The effect of capacity is directly opposite to self- 
induction. Hence by properly arranging the capacity 
of a circuit it is possible to neutralize inductance, and so 
to bring the alternating current under the same laws as 
direct. If /be the frequency and E c the pressure at any 
moment applied to a condenser, or circuit whose con- 
denser capacity is C, the current flowing back and forth 
due to this capacity is 

I=2TrfCE c . (126) 

and E < = 7^fC' (I2? ) 

In this case the resistance due to capacity is 

E c may be called the capacity pressure which is opposite to 



238 ELECTRICAL AND MAGNETIC CALCULATIONS. 

JS sy the self-induction pressure. Fig. 25 represents the re- 
lation of capacity to other elements of an alternating cur- 
rent circuit. 



£~zwjo 




(/) The ratio of the inductance L of an alternating 
current circuit to its resistance R is called the time con- 
stant of the circuit. 

R m ( I2 9) 



Time constant 



" The value of the time constant is a measure of the growth 
of the current " in an inductive circuit. When no induc- 
tance is present the current instantly reaches the value 

— in which E is the impressed electromotive force, and 
R 

R the resistance of the circuit. Otherwise it takes time 

L 



for the maximum value to be reached proportional to 



R 



But if the resistance is increased in proportion to L the 
time may be the same for large as for small inductances. 

60. Inductance, Capacity and Resistance. Series. — Ex- 
ample. — How many volts must an alternator with disk 
armature without iron generate when it is desired to have 
1000 terminal voltage, the current being 25 amperes, fre- 



ALTERXATIXG CURRENTS. 239 

quency 100 cycles per second, resistance 1 ohm, and 
coefficient of inductance L = 0.01 henry ? — Jackson.* 

Solution. — Reference to Fig. 24 and formula (120; will 
give for the E.M.F. of self-induction 

E s = 2-jt/LI = 27r X 100X0.01 x 25 = 157 volts. 

The active pressure from (122) will be 

E a = IR = 25 x 1 = 25 volts, 

The external required voltage is 1000 volts. Therefore 
the total active pressure becomes 

E a = 25 + 1000 =1025 volts. 

Hence the impressed pressure to be developed by the 
alternator is from (123) 

E { = \IE?+E a *= V^+T^ 2 = I037 volts. 

Example. — How many more volts than in the preced- 
ing case must be generated in the armature of the above 
machine if iron be used in the core so that when carry- 
ing 25 amperes the inductance is 0.025 henry ? 

Solution. — As before the total active pressure is 
E a = 1000 + 25 = 1025 volts, 
and E s = 27r x 100 x 0.025 x 2 5 = 39 2 -7 volts. 

Therefore the total impressed E.M.F. must be 

2 s 

1025" + 392.7" = 1098 volts, 
or 1098 — 1037 = 61 volts higher pressure than before. 

When two or more inductances are in series, and have 
equal time constants, that is, if 

E 1 R 2 R z 

* See Jackson's A Iternating Cnrrenti, p. 75, 



240 ELECTRICAL AND MAGNETIC CALCULATIONS. 

then the total reactance is the sum of the individual re- 
actances, and the total drop of potential over them is the 
sum of the individual drops. If the time constants are 
not equal the resultant impedance and potential must be 
found in the third side of the triangle whose other two 
sides are the separate impedances and potentials. 

Example. — A circuit has a resistance of 10 ohms and 
a capacity of ioo microfarads = 0.000100 farad. What 
must be the impressed E.M.F. at a frequency of 127 \ to 
send 10 amperes through the circuit? How many am- 
peres will flow if the impressed pressure be 160 volts? 
What is the angle of lead ? 



R-10 



A/WWV^ 



C = 100 

Fig. 26. 



Solution. — The relation may be indicated as in Fig. 26, 
in which C is the condenser, and R the resistance. Also 
referring to Fig. 25 and formula (127), 

E c = -^ = = 125 volts. 

2 7T/C 2 7T X I27.5 X 0.000I00 

E n = io X 10 = 100 volts. 



Therefore, E { = v ioo 2 + 125 2 = 160 volts. 

Impedance = V7o + 12.5 = 16 ohms. 

Also /at 160 volts is 

r E 160 

I = — = —t- = 10 amperes. 
K 16 



ALTERNATING CURRENTS. 24 1 

The angle of lead, or negative lag, is the angle whose 

tangent is 

10 

tan d> = = 0.8. 

12.5 

The angle corresponding to this tangent from a table of 
natural functions is (table 14) 

<£=- 51° 20'. 

Example. — Suppose a circuit to have a negligible 
ohmic resistance, a capacity of 100 microfarads and an 
inductance of 0.0 1 henry in series ; if the frequency is 
taken at 127^- for convenience in calculation, how many 
volts must be applied to send 10 amperes through the 
circuit? If 220 volts are impressed upon the circuit how 
many amperes will flow ? 

Solution. — It has already been shown that capacity 
and inductance act oppositely or 180 apart. Reference 
to Figs. 24 and 25 shows how they are represented dia- 
gramatically. We may then find the reactance due to 



C ° 100 L-JD1 



sWMP 



Fig. 27. 

each, and plat by oppositely directed vertical lines — the 
inductive resistance upward from a point, and the capa- 
city resistance downward from the same point. The 
difference is then the resultant reactance. 
From (121) 

R 8 = 2-rrfL = 2 7r X 127^- x 0.01 = 8 ohms. 



242 ELECTRICAL AND MAGNETIC CALCULATIONS. 

Also from (128) 

R c = — -r^ = -. = 12.5 ohms. 

2 TTjC 2 7T X I27-2- X O.OOOI 

Therefore the resultant reactance is 

R = 12.5 — 8 = 4.5 ohms. 

Hence from Ohm's law 

E = IR = 10 X 4.5 =45 volts. 

Also at 220 volts the current would be 

E 220 
7 = — = = 48.8 amperes. 

-«■ 4-5 

It will be obvious from this that it is possible so to adjust 
a capacity that its resistance will be equal to the induc- 
tive resistance, and thus counteract it and make the 
circuit equivalent to one with neither inductance nor 
capacity. 

When two capacities are in series the total resistance 
is the sum of the two resistances and the total fall of 
potential is the sum of the individual drops. 

Example. — Derive the formula for calculating the 
E.M.F. of self-induction in a coil of known constants. 

Solution. — The E.M.F. in any coil is proportional 
to the rate of change of the lines of force threading 
through the coil, and to the number of turns in the coil. 
If N be the whole change in the number of lines of force 
due to any cause in ^seconds, and the number of turns 
is represented by n, the E.M.F. will be 



ALTERNATING CURRENTS. 243 

nN 

In this equation — r gives absolute units of E.M.F., or 

dynes, and io 8 reduces to practical units of E.M.F. In a 
long solenoid 

N=±——> ( I3I ) 



in which ;/ is the number of turns per centimeter of 

length of the solenoid, A the area of the coil, and — 

absolute amperes, I being practical amperes. Substitut- 
ing this value of N in (130) gives for the self-induced 
electromotive force 

\TT?l?l'Al 



The number of lines passing through the coil when the 
current is 1 C.G.S. unit is \ir?i f A, If the coil have 
iron in it so that its permeability becomes /x instead of 
1, (132) will become 

Also in (132) ^ir7i?i r A = L, the coefficie?it of self-i?idiutio?i, 
and we may write 

Here L is expressed in absolute units. If L is given in 
henrys, 

E = -T' ( j 35) 



244 ELECTRICAL AND MAGNETIC CALCULATIONS. 

When (133) is applicable, L = ^.irnn'Aix absolute units ; or 

4 irnn'Au. , 
L = ^ — nenrys. 

io a J 

This again gives (135) 

e = lI. 

Therefore the E.M.F. of self-induction is equal to the co- 
efficient of self-induction in henrys multiplied by the rate of 
change of the current in amperes per second. 

Since 

N = — , and E = n 



10 io 9 r io 9 ! 7 

therefore we may express, for unit current and permeability, 

z ^7^- (^36) 

Example. — How many volts of counter E.M.F. will 
be developed in a solenoid 50 centimeters long uniformly 
wound with 250 turns of wire, the area being 4 square 
centimeters, and in which the current of 5 amperes takes 
0.0 1 second to rise from zero to its maximum value ? 

Solution. — n= 250 turns, /= 50 cm.; hence n r = 
250 -f- 50 = 5 turns per cm. Also A = 4 sq. cm., /x = 1, 
and T= y l-^ second. Applying (133) 

47T X 250 X5X4X1X5 r . 

^ = — ^— ~ = 0.00628 volt. 

io y X 0.01 

Example. — Suppose that in the above example a core 
of iron whose permeability may be taken to be 500 were 
put in the solenoid How many volts of counter E.M.F. 
will be set up in the coil ? 



ALTERNATING CURRENTS. 245 

4^X250X5X4X500 

Solution. — E =- ^ — - — - — - — = 3.14 volts. 

io 9 x 0.01 ° 

Example. — Find the coefficient of self-induction of a 
coil of 250 turns uniformly wound on an iron ring 100 
centimeters in mean circumference and having a cross 
sectional area of 20 square centimeters, and carrying a 
current varying at the rate of 2 amperes per second, p 
being 250. — Jackson. 

Solution. — Since N = > 

10 

10 X N , _ 

— = 47172 AfjL = 47r X 25 X 20 X 250 = 1,571,000, 

the number of lines per ampere per second. Therefore 

nN 1,01,000 X 2croo , 

L = — = -^ 5 ^— = 3-93 henrys. 

io y io 9 

Otherwise, since \Trnn r Aix.I 

^ = 9 ' 

io y 

where / is the number of amperes per second, and since 

L is measured by the number of volts of counter E.M.F. 

when the current changes at the rate of 1 ampere per second, 

4.7rnn'AfjL 4^X2500X25X20X250 

£=lz = — = 

I o y I o y 

= 3.93 henry s, as before. 

61. Inductance, Capacity, and Resistance. Parallel. — 

Example. — What is the resistance when 5 ohms and 
1 5 ohms are connected in parallel ? How many volts 
will cause 20 amperes to flow ? 

Solution. — From (17) 

R R^ H t S iS iS 



246 ELECTRICAL AND MAGNETIC CALCULATIONS. 

Hence 

R = J ? 5_ = 2.75 ohms, and E = IR = 20 X 3.75 = 75 volts. 

Example. — Suppose two inductances with negligible 
resistances^! = 0.0 1 and Z 2 = 0.05, be placed in parallel 
on a circuit whose frequency is 63.75. Find the joint 
impedance and the current when the voltage is 200.. 

Solution. — Since there are no ohmic resistances the 
time constants are equal and the same rule applies as in 
the last example ; namely, the combined impedance is 
the reciprocal of the sum of the reciprocals of the 
individual reactances. Hence 

Z = ^^ + 2^7^' ( I37 ^ 

Whence 

1 1 1 6 

R ~ 2 7T X 63.75 X °* 01 2 7T X 63.75 X O.O5 " " 20 

Therefore 20 , . 

R = — = 3 J ohms, 

and T E 200 r 

/ = — = — - = 60 amperes. 

Example. — An inductance of 0.02 henry is connected 
in parallel with a resistance of 20 ohms. What is the 
impedance, and how many volts are required for 50 
amperes, when the frequency is 78.6 so as to make 

27l/= 5OO ? 

Solution. — The connections are indicated in Fig. 28. 
The time constants are not alike, hence we must take 
the geometric sum of the reciprocals as the reciprocal 



ALTERNATING CURRENTS. 



247 



of the required impedance. That is, the combined con- 
ductivity will be the hypotenuse of the right triangle of 



R=20 



r^WVWWWK 



L=.02 



sMMM^y 



Pig. 28. 

which the ohmic conductivity and the reactive conduc 
tivity are the two sides, respectively. 

11 , 1 1 

Tr = — = 0.05, and — = — = 0.10. 

R^ 20 °' 2 irfL 10 

These values are then represented as in Fig. 29. 



Ri = 


20 


\ v 


05 






% 






V 



Fig. ag, 



27T/L 



rW=- 



248 ELECTRICAL AND MAGNETIC CALCULATIONS. 



Whence Impedance R = 



0.1 1.1 



= 9 ohms, 



and 



E = IR = 50 X 9 = 450 volts. 



Example. — If a resistance of i.6f ohms be placed in 
parallel with a capacity of 1000 microfarads and a pres- 
sure of 100 volts applied at a frequency of 127^, how 
many amperes will flow in the circuit ? 

Solution. — This is similar to the preceding and 
Fig. 30 shows the connections. We must again repre- 



HWWW\AA 



R,-Wy, 



C=1000 



Fig. 30. 




.8=27T/C 



sent the conductances by the two sides of a right triangle ; 
the hypotenuse will be the total conductance whose recip- 
rocal will be the required impedance. 

i=^k = °- 6 ' 

and 2 7rfC = 2 7rX 127^- X 0.001000 = 0.8. 



Whence 
Therefore 

Hence 



1 = V0.6 2 + 0.8 2 = 1, and R = 1, 
R 

E 100 

I = — = =100 amperes. 

R 1 

rr, O.8 

Tangent <j> = — - = 1.333. 
0.0 

<£ = 53 8", lead angle. 



ALTERNATING CURRENTS. 249 

Example. — A capacity of 50 microfarads is connected 
in parallel with an inductance of 0.05 henry; when the 
frequency is 127.5 how many volts will be required for 
10 amperes ? 

Solution. — Capacity and inductance lines are opposed 
to each other and are drawn vertically. Fig. 32 gives 
the connections. The conductances are 

2 tt/C = 0.04, and — — = 0.025, 

2 TTjL 

Therefore — =^o.o4 2 + 0.025 = 0.0149, 

R 

and R = = 67 ohms. 

0.0149 

Also E = IR = 10 x 67 = 670 volts. 

Example. — There are in parallel two circuits as 
follows: the first having R x = 8 ohms and L x = 0.0075 
henry; the second having R 2 = 12.5 ohms and Z 2 = 
0.0125 ; if the frequency is 127.5, wna -t is the total im- 
pedance? 

^ = 8 : L,-.0075 




12.5: U -\01 25 



Fig. 33. 

Solution. — Connections are represented in Fig. 330 
First find 

R' = V^2 + Y^r/% 2 = Vs 2 + 6 2 = 10 ohms. 

R"= ^R 2 2 + IrfZ? = ^7^5 2 + io 2 = 16 ohms. 



25O ELECTRICAL AND MAGNETIC CALCULATIONS. 



These are obtained graphically as in Fig. 34. The con- 
ductances are -X- = 0.1, and -J* = 0.0625 respectively. 
Now represent these on some convenient scale, say one 
inch for 0.05, or 2 inches for 0.1, and 'draw them at 
the proper angles as in Fig. 35. Lay down OL hori- 



R 2 =12.5 




27T/Lr 10 




Fig. 34. 



Fig. 35. 



zontally, then 0(9=0.1, making the same angle with 
<9Z, as OG in Fig. 34. From G draw GK in the same 
direction as OK and make it equal to 0.0625. Then 
draw OK and measure it ; then from the scale taken 
determine its length. We find it to be 0.1625. The 
impedance is its reciprocal, or 6.15 ohms. 

Example. — A non-inductive resistance of 5 ohms is in 
series with an inductive resistance of 5 ohms and 0.02 
henry. What impressed pressure at a frequency of 100 is 
necessary to furnish 2 5 amperes ? 

Solution. — As before find the impedance of the in- 
ductive path. Both from Fig. 36 and (123), 

R= v^ 2 2 + 2 ?r/Z 2 = ^5 2 + 2 7T X 100 X o.o2 2 = 13.57 ohms. 



ALTERNATING CURRENTS. 



251 



The conductance of the reactive branch is then 



!3-57 



27?/L = 12.56 



= 0.0736, which is platted to scale in Fig. 2>h having the 
same direction as LK in Fig. 36. ^ r,=5 
From K in Fig. 37 KP is 
drawn = \ = 0.2. the conduc- 
tance of the non-inductive 
branch. LP is then drawn and 
measured, and from the scale 
chosen is found to be 0.261, the 
total conductance. The recip- 
rocal gives the impedance = 
3.83 ohms. This is also ob- 
tained from the trigonometric 
relations, 
will be 




Fig. 36. 

Hence for 25 amperes the pressure in volts 



E = 25 X z&z = 95.75 volts. 



O'— 



^7 







K 


.20 




Fig. 37. 



62. Problems. — 1. Find the average and the effective 
E.M.F. of an alternator whose maximum E.M.F. is known 
from its curve to be 1500 volts. 



Average = 955.5 volts. 
Effective = 1060.5 volt. 



252 ELECTRICAL AND MAGNETIC CALCULATIONS. 

2. If a flat coil of wire develops 5 volts when its plane 
is parallel with the lines of force of the magnetic field, 
what will be the E.M.F. at the instant when its plane 
stands at an angle of 45 with respect to the lines of 
force ? 

Required E\ 5 volts : : sin 45 : sin 90 , or 
E 15 volts : : \ V2 : 1 

E = f V2 = 3.5 volts. 

3. How many amperes of current will flow in a circuit 
under a dynamo pressure of 1200 volts, when the com- 
bined resistance of all the circuit is 20 ohms and the 
average coefficient of self-induction is 0.0312 henry, fre- 
quency 127.5? ^ == 37-S amperes. 

4. A dynamo having 8 poles makes 2200 r.p.m. It 
has a resistance of 0.1 ohm and an inductance of 0.01 
henry. What E.M.F. must it generate to give a terminal 
pressure of 1000 volts if the current is 25 amperes? 
Draw proper diagrams. 

E s = 2 tt/LI — 230 volts. 

E a = Z ff = 2.5 volts. 

Therefore, E { = V(iooo + 2.5)2 + (230) 2 
= 1028.6 volts. 

5. If an 8-pole machine is to run at 1800 r.p.m and 
has a coefficient of self-induction of 0.005 henry, resis- 
tance of 2 ohms, what voltage must be generated in the 
armature winding so that a terminal voltage of 1000 may 
be furnished when the current is 25 amperes? 

E i = 1054.2 volts. 



ALTERNATING CURRENTS. 253 

6. The pressure applied to a circuit whose resistance 
was 5 ohms was 250.32 volts, the circuit carrying 50 am- 
peres ; what was the E.M.F. of self-induction ? Also if 
the frequency was 63.75, what was the coefficient of 
inductance? E s = 12.65 volts. 

L = 0.00063 henry. 

7. How many amperes will pass through a circuit 
under 100.7 volts applied at a frequency of 63.75, wnen 
the resistance is 10 ohms and inductance 0.0 1 henry? 
What are the ohmic and inductive drops? 

/ = 10 amperes. 
E a = 100 yolts. 
E $ = 40 volts. 

8. What resistance may be added in a circuit whose 
inductive drop at 100 amperes is 300 volts, when 500 
volts is the impressed pressure ? What is the active pres- 
sure and impedance ? R x = 4 ohms. 

E a = 400 volts. 
R = 5 ohms. 

9. What E.M.F. at a frequency of 125 must be applied 
to a condenser circuit whose capacity is 100 microfarads, 
so that 1 o amperes of current will flow ? 

E { = 127.4 volts. 

10. What capacity must be put into an alternating cur- 
rent circuit of negligible resistance, so that 50 amperes 
may be obtained at a pressure of 500 volts and frequency 
of 1 27 J? C , = i25 microfarads. 

1 1 . What is the capacity reactance in problem 10? 

R = 0.1 ohm. 



2 54 ELECTRICAL AND MAGNETIC CALCULATIONS. 

12. What is the impedance of a circuit which contains 
an ohmic resistance of 40 ohms in series with a capacity 
resistance of 30 ohms ? 

R = V 30 2 + 40 2 = 50 ohms. 

13. If 10 amperes are required through the circuit in 
problem 12, how many volts of E.M.F. must be impressed 
upon it ? Find the ohmic and capacity drops. 

Ei = 10 X 50 = 500 volts. 
E c = 10 X 30 = 300 volts. 
E a = 10 x 40 = 400 volts. 

14. A dynamo supplies 25 amperes to a circuit con- 
sisting of 10 ohms in series with a capacity of 100 micro- 
farads at a frequency of 63.75. What is the terminal 
voltage of the dynamo, and what is the drop in each por- 
tion of the circuit? E { • = 673 volts. 

E a = 250 volts. 
E c = 625 volts. 

15. When a capacity of 10 microfarads is in series 
with a resistance of 10 ohms, and an E.M.F. of 638 volts 
at a frequency of 100 is impressed upon the circuit, how 
many amperes will pass through it, and what will be the 
drop of potential on each portion of the circuit ? 

1=4 amperes. 
E a = 40 volts. 
E c = 636.8 volts. 

16. What is the time constant of a circuit whose resist- 
ance is 10 ohms and whose inductance is 0.05 henry? 

L o.oq 

/ = — = =0.005 sec# 

K 10 



ALTERNATING CURRENTS. 255 

17. What is the coefficient of self-induction of a circuit 
whose time constant is 0.0 1 second and whose resistance 
is 5 ohms. L = 0.05 henry. 

18. An alternating current circuit is carrying 5 am- 
peres under an impressed pressure of 50 volts at a 
frequency of 1271. If its resistance is 6 ohms, w r hat is 
its time constant ? 

/ = — = 0.00 1 6t seconds. 
R 3 

19. What current will flow through a circuit having a 
negligible resistance, an inductance of 0.05 henry, and 
a capacity of 1 microfarad in series, when a pressure of 
500 volts is applied at a frequency of 100 ? Draw 
diagram. /= 0.32 ampere. 

20. What capacity must be put in series with an in- 
ductance of 0.02 henry when the • frequency is 63.75 so 
that the impedance shall be zero ? 

(7=312 microfarads. 

21. How many amperes will flow under the conditions 
given in problem 20, if a resistance of 50 ohms be added 
in series and 1000 volts E.M.F. be applied? 

I = 20 amperes. 

22. What is the coefficient of self-induction of a coil of 
500 turns through which the magnetism is changing at 
the rate of io 5 lines of force per second ? 

L = nlV-r- io 9 = 0.05 henry. 

23. How many lines of force are produced in a coil 
having 10 turns per centimeter of its length and a cross 



256 ELECTRICAL AND MAGNETIC CALCULATIONS. 

section of 10 square centimeters when it carries 10 am- 
peres of current, no iron being in its vicinity ? 

n _ \TrriAI r .. 

JV= = 1256 lines. 

10 ° 

24. How would the number of lines be affected if an 
iron core were introduced into the coil, and its permeability- 
be given as //, = 500 ? 

4 TTfl A Iu. - _ 

JV= =628,000 lines. 

10 

25. If the total length of the coil in problem 24 is 50 
centimeters, what E.M.F. of self-induction is set up under 
the conditions named ? 

E = ± g-t- = 3.14 volt. 

io y 

26. What is the inductance under the conditions 
named in problem 25 ? 

L == '-= = in absolute value. 

E ATrnn'Au, 
L= -j = —^— = 0.314 henry, 

in practical units. 

27. If a voltage of 1040 at a frequency of 125 be ap- 
plied to a circuit consisting of a resistance of 10 ohms, an 
inductance of 0.05 henry, and a capacity of 100 micro- 
farads, all in series, how many amperes of current will 
flow in the circuit, and what is the impedance ? Also the 
drop over each portion of the circuit ? 

/ = 36.7 amperes. 
R = 28.3 ohms. 
E a = 367 volts, 
E c = 466 volts. 



ALTERNATING CURRENTS. 2 $7 

28. There are in parallel two inductances, I^= 0.0002 
henry, Z 2 = 0.0004 henry. The frequency is 100. What 
is the impedance, and how much current will flow under 
an impressed pressure of 8.37 volts ? 

R = 0.0837 onm - 
/ = 100 amperes. 

29. An inductance Z = o.o5 henry is placed in parallel 
with a non-inductive resistance of 10 ohms. When a volt- 
age of 177 is applied to the circuit at a frequency of 60, 
how many amperes will flow ? R = 8.85 ohms. 

I = 20 amperes. 

30. Let an inductance of 0.0 1 henry be placed in 
parallel with a capacity of 50 microfarads; what E.M.F. 
at a frequency of 127^- must be applied to the circuit to 
give 10 amperes of current ? How many amperes will 
pass in each branch ? R = 11.77 ohms. 

E = 1 1 7.7 volts. 
I s = 14.71 amperes. 
I c =4.71 amperes. 

31. A coil of wire has an ohmic resistance of 10 ohms 
and an inductance of 0.0156 henry. What is the im- 
pedance when the frequency is 127J? 

R = 16 ohms. 

32. A capacity circuit has a capacity of 156^ micro- 
farads in series with a resistance of 10 ohms. Find the 
impedance at a frequency of 127 J. R = 12.8 ohms. 

^^. Place the circuit of problem 31 in parallel with that 
of problem 32 and find the joint impedance. Find the 



258 ELECTRICAL AND MAGNETIC CALCULATIONS. 



number of amperes of current when 787 volts of E.M.F. 
are applied. R = 7.87 ohms. 

/ = 100 amperes. 

<£ = 19 42'. 

Solution. — R s = ^io 2 + i2»5 2 = 16 ohms. 
R c = ^io 2 + 8 2 = 12.8 ohms. 

i = ie = °- o62S - 

±. = -i- = o.o 7 8. 

In the second diagram, Fig. 38, plat conductances 
0.078 and 0.0625 Parallel to their respective resistances, 
then complete the triangle whose third side will to scale 
represent the total conductance. Hence 



Whence 



R 

R = 



= 0.127, 

1 



0.127 



= 7.87 ohms. 



27T/L- 12.5 





Fig. 38. 

34. An inductive resistance of 8 ohms and coefficient 
of self-induction of 0.015 henry is connected in parallel 



ALTERNATING CURRENTS. 



259 



with a circuit having 8 ohms resistance and 4i6f micro- 
farads capacity. Find the joint impedance and the 
current flowing when the impressed E.M.F. is 1350 volts, 
the frequency being 63.75. 

R = 6.25 ohms. 

7 = 200 amperes. 

<t> = o°. 



Solution. — ^, = 10 is the impedance of the first 
circuit, R c = 10 is the impedance of the second. The 
conductances corresponding are each 0.1. These are 
represented to scale by A L and L B, Fig. 39, drawn to 

C 



2TT/L--6 




r=6 



2 7T/C 



Fig. 39. 

each other at the same angle as A D and A C. The 
resultant is A B with no lag. A B = 0.1599, tne recip- 
rocal of which is 6.25 ohms, the desired total impedance. 



Therefore 



E 1350 
R 6.25 



200 amperes, 



26o ELECTRICAL AND MAGNETIC CALCULATIONS. 



35. Find the joint impedance when an inductance of 
0.02 henry is in parallel with a capacity of 312.5 micro- 
farads, the ohmic resistance being negligible in each case, 
and the frequency is 63.75. ' i? = infinity. 

36. How much current would flow in the circuit if 100 
ohms were put in parallel with the combination in prob- 
lem 35, and 1000 volts be applied ? f=*ffip-= 10 amps. 

37. A non-inductive resistance of 50 ohms is in parallel 
with a capacity of 200 microfarads. Find the joint impe- 
dance. Also now place an inductance of 0.015 henry in 
series with the parallel combination, and find the total 
impedance and the E.M.F. necessary for 10 amperes 
when the frequency is given as 100. R c = 7.87 ohms. 

jR = 2.08 ohms. <£ = 7°28 / . E = 20.8 volts. 

Solution. — Joint im- 
pedance of parallel part is 
obtained by finding the 
third side of the triangle. 
1 

R 

base and 2 ir/C = 6.28 X 
100 x 0.0002 = 0.1256 
forming the perpendicular. 
The hypotenuse is 0.127. 
¥i ^- 40. Its reciprocal = 7.87 ohms. 

Now lay the latter down to scale and at the same angle 
with the horizontal as O L and from L draw upwards 
to scale LK '= 2-nfL = 6.28 x 100 X 0.015 = 9.4 ohms. 
Then OK = 2.08, the joint impedance ; <j> = 7°28'= angle 
lag. 




2TTfc=A256 




= — = 0.02 forming 



ALTERNATING CURRENT DISTRIBUTION 26 1 



XIII. 
ALTERNATING CURRENT DISTRIBUTION. 

63. Alternating Current Circuits. — It has already been 
shown that there are several causes making it inaccurate 
to apply Ohm's law strictly to calculations in alternating 
currents. However, where the transformer secondaries 
are loaded with incandescent lamps, the effect of induc- 
tance becomes negligibly small, so that the inaccuracies 
will not be very great for moderate lengths of circuits. 

Practically the ratio of transformation of transformers 
is the ratio of the primary turns to the secondary turns, or 
the ratio of the secondary current to the primary current. 
The ratio of transformation of transformers used for in- 
candescent lighting on single-phase circuits is usually 20 or 
10. In the monocyclic * system ratios of 9 and 4 are often 
used. In long distance polyphase transmission various 
ratios are employed, particularly in the large sizes of 
transformers. In a town having a population of 5000 to 
10,000, the lighting plant being centrally located, the 
primary E.M.F. will be, perhaps, 2000 to 2200 volts, and 
the secondary 100 to no volts, the ratio being 20. Where 
the primary mains are quite short, possibly the primary 
E.M.F. would be 1000 volts, and the ratio of transformation 
10. In the Niagara-Buffalo line, the power is brought to 
the city limits at 22,000 volts, then transformed to 11,000, 
the ratio of transformation being in this case but 2. 

* The monocyclic system is no longer a commercial system. 



262 ELECTRICAL AND MAGNETIC CALCULATIONS. 

The efficiency of the larger sizes of transformers is from 
95 per cent to 99 per cent, and the core losses and the 
copper losses are about equal. To obtain a given energy 
in the secondary, the primary current will be increased by 
the amount of the core loss per cent, while the primary 
voltage will be increased by the per cent of copper loss. 

Example. — When the secondary of a transformer is 
wound for 52 volts, and the ratio of transformation is 20, 
what is the primary current for a 1000 watt transformer? 

Solution. — Secondary current 

I 2 = 1000 -*- 50 = 20 amperes. 

This will furnish 20 lights, 16 c.p. 

Primary voltage = 52 X 20 = 1040 volts. 
Primary current 7j= 20 -1- 20 = 1 ampere. 
Therefore, 

Watts in primary for no loss = 1040 X 1 = 1040 watts. 

If the loss be 5 per cent, then 

Watts in primary coil = 1040 -7- 0.95 = 1095. 

Suppose copper and iron losses equal, each being 2\ per 
cent. Then the primary current is 

I x = i-r- 0.975 = 1.025 amperes. 
Primary E.M.F. 

E x = 1040 -7- 0.975 = 1066.66 volts. 

Otherwise, E x = 1095 -f- 1.025 = io ^7 v °lts. 

Strictly, the ratio of transformation here is 20 -f- 1.025 = 
19.4 instead of 20. Hence the transformer * losses have 
the effect of reducing the ratio of transformation. 

* This refers to the iron losses. 



ALTERNATING CURRENT DISTRIBUTION 263 

Example. — What is the maximum per cent drop in the 
secondaries in the last example if 50 volt, 1 ampere 
lamps are to be used ? What size of wire will be used, and 
what must be the size of the primary mains, neglecting 
drop due to inductance, if an ohmic loss of 10 per cent be 
allowed, and the distance be \ mile ? 

Solution. — Secondary loss = 2 volts -2-52 =3.8 per 
cent. Assume the distance to the lamps to be 100 feet. 
Then the line resistance per 1000 feet is 

R 9 = (2 -r- 20) X = o.q ohm, which is No. 7 A.W.G. 

2 v J 200 ° ' 

For the primary, I x = 1.025 amperes ; 

E (drop) = — 1067 = 110 volts. 

v ' 0.90 

Hence R x = 119 -*- 1.025 = 116 ohms for 5280 feet. 

Making per 1000 feet, R x = 116 -f- 5.28 = 22 ohms, 
or No. 23 wire. This loss, of course, is excessive. 

Machine voltage = 1067 -*- 0.90 = 1186 volts. 

The table No. 10 will give the impedance, the ohmic re- 
sistance and the inductive resistance in ohms per mile of 
those sizes in which there will be an appreciable effect at 
the given distances of the wires apart. These values are 
given for frequencies of 60 and 125 per second. The 
column headed R x means ohms resistance per mile, R s is 
the inductive resistance in ohms per mile, and R is the 
impedance or resultant resistance in ohms per mile. It 
will be observed that the values in the columns headed R 
are obtained from 

R = \lR? + 4 tt 2 / 2 L 2 = \IR* + R 9 \ 



264 ELECTRICAL AND MAGNETIC CALCULATIONS. 

Thus for No. o at 12 inches apart, 

R = ^0.519* + ^55 2 = 0.756. 

It is safer to add 15% to the tabular inductive resis- 
tances to compensate for distortion from the sine wave. 

Example. — By means of the table determine the 
E.M.F. necessary to apply to a circuit \\ miles long to 
carry 10 amperes, when No. 1 wire is used 12 inches 
apart, and the frequency is 60. 

Solution. — 

R x of No. 1 for 3 miles = 0.655 x 3 = x -965 ohms. 

R s of No. 1, 3 miles = 0.565 X 3 = 1.695 ohms. 

Impedance R = 0.865 X 3 = 2.595 ohms, 

or R = V 1. 695 s + 1.965 2 = 2.595 ohms. 

Since E = IR, 1= 10, and R = 2.595, 

£=ioX 2 -S9S = 2 5«95 volts line drop. 
If there were no self-induction E would be 
1.965 X 10 =19.65 volts drop. 

Therefore 25.95 — 19-65 = 6.30 volts necessary for 
inductive drop on line. 

At 125 cycles per second this would be 

E = 1.349 X 3 X 10 = 40.47 volts total drop. 

Inductive drop is 40.47 — 19.65 = 20.82 volts. If the 
load is non-inductive, as incandescent lamps, the gener- 
ator must furnish 20.82 volts more on account of line 
self-induction. 



ALTERNATING CURRENT DISTRIBUTION 265 



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266 ELECTRICAL AND MAGNETIC CALCULATIONS. 

Let the load be 20 groups of no volt incandescent 
lamps, 5 in series in each group. This will require 
110X5=550 volts at the lamp terminals. The resis- 

c X 220 

tance of the load is therefore = c; c ohms. This 

20 °° 

is in series with the line resistance and inductance. 
Take the frequency at 60, then neglecting dynamo resis- 
tance and inductance, the impedance of the circuit 
will be 

R = Vi. 9 6 5 + 55 2 + 1.695 2 = 57 ohms. 

The total impressed E.M.F. must be 

E = IR = 10 X 57 = 570 volts. 

Hence the line drop due to all causes is 570 — 550 = 
20 volts, while the ohmic drop alone is 19.65 volts. 
Therefore it is plain that for non-inductive loads and very 
small machine inductance, the line self-induction is not 
a very disturbing factor. 

Example. — Assume the machine resistance to be 
0.4 ohm and the coefficient of self-induction L = 0.02 
henry, speed so as to make / = 60 cycles, supplying 
10 amperes as the above load. Find the total impedance 
and the E.M.F. in the armature conductors. 

Solution. — Total reactance is ^=1.695 -f- 27r X 60 X 
0.02 = 9.231 ohms. The total impedance is 

R = ^56.965 2 + 9.231 2 = 58 ohms. 

This is not appreciably different from that found above 
because the resistance is quite large in proportion to the 



ALTERNATING CURRENT DISTRIBUTION. 267 

reactance ; that is, the time constant is very small. Hence 
the armature self-induction has not seriously disturbed 
the regulation. 

Example. — It is required to determine the size of 
wire at 12 inches apart for the circuit and load given 
above so that the total drop shall not exceed 25 volts. 

Solution. — E.M.F. at load =550 volts. 
Impressed E.M.F. = 550 + 25 = 575 volts. 
Armature reactance R s = 7.536 ohms = 27r x 60 x 0.02. 
Let the line resistance be R v then the impedance 



R = Ytf^ + x + 7 . 53 6 2 = -Vo 5 - = 57-5 ohms. 

Solving for R x and neglecting terms containing x, since 
the line reactance is relatively small, we get 

_R X = 57.4 ohms. 
Therefore the line resistance per mile is 

57 ' 4 "" 55 = 0.823 ohm. 

According to the table this corresponds to No. 2 wire. 

Taking into account the transformer inductance and 
the transformer and lamp losses under ordinary condi- 
tions, when the transformers cannot be counted on as 
more than three-fourths loaded, it may be assumed with- 
out great error that the self-induction will cause a lag of 
8° or less, making the power factor 0.99 and the induc- 
tance factor 0.14. In case of very light loads on trans- 
former secondaries, the figures may become in practice 
0.98 for the power factor and 0.19 for the inductance 
factor, and u° for the angle of lag. 



268 ELECTRICAL AND MAGNETIC CALCULATIONS. 

Where transformers supply motor loads the power 
factor would probably run down to 75% or 80%. How- 
ever, if both transformers and motors could be fully loaded 
— a rare thing in practice — the power factor with mod- 
ern appliances might go over 90%. 

Example. — Taking a circuit whose power factor is 
99%) what is the proportion of impressed E.M.F. to 
active E.M.F. in the circuit? 



Solution. — 

tanrp wnn 4- ta too 

approximately. 



E { Impedance V99 2 + 14 2 100 



E a Resistance 99 99 

In other words, when the E.M.F. necessary to force the 
desired current through the given resistance would be 
990 volts, the impressed voltage must be 1000, 10 volts 
being required to overcome the inductive resistance. 

Example. — Taking 0.80 as an average power factor 
for motor loads, the inductance factor will be about 0.595 
and the lag 36I- . Find the impedance factor and ratio 
of impressed volts to active E.M.F. 

Solution. — 

Impedance = ^o.8o 2 + 0.595 2 = 0.997 ohm. 

E { 0.997 



E„ 0.80 



= 1.246. 



In words, if 1000 volts would send the desired current 
through the given resistance, 1246 volts must be im- 
pressed upon the circuit; 1246 — 1000 = 246 volts extra 
are required, due to self-induction. 



ALTERNATING CURRENT DISTRIBUTION. 269 

The following examples have been taken from Emmet's 
" Alternating Current Wiring and Distribution." They 
will serve to illustrate the type of solution which may be 
applied to problems under the present section. In such 
problems as these that follow consider the ratio of trans- 
formation one, and use the primary voltage and current 
on both sides of the transformer. 

Example. — Assume 500 incandescent lamps of 57.5 
watts each on secondaries of transformers of different 
sizes, half loaded, on the average. The mains to the 
transformers consist of No. 2 B. & S. wire 2 miles long 
and 18 inches apart. The frequency is 125. The lamp 
voltage is to be 100 and the ratio of transformation is 10. 
Determine the voltage required at the generator and the 
ohmic drop on the line. 

Solution. — 

Lamp active volts = 1000 ; lamp current 28.75. 

Secondary ohmic drop 3% = 30 volts; inductive drop 
3% = 30 volts. 

Transformers, ohmic drop 1% = 10 volts ; inductive 
drop i2^% = 130 volts. 

Core loss 5% of 28.75 = x -44 amperes. Total / = 
30.19. 

Line current = 30.19 amperes. Ohmic drop = (o.826 X 
4) X 30.19 = 99 volts; inductive drop (1.31 X 4) X 
30.19 + 15% = 182 volts. 

Now add active volts and ohmic drops for total active 
volts. 



270 ELECTRICAL AND MAGNETIC CALCULATIONS. 

Whence E a = iooo + 30 + 10 + 99 = 1139 v °lts. 

Also add inductive drops giving 

E s = 30 + 130 + 182 = 342 volts. 

Impressed pressure 

E i = V1139 2 + 342 2 = 1 188 volts. 

The line loss due to ohmic resistance = 99 volts. 

Total line loss = 30 + 10 + 99 = 139 volts ; 
or II 39 — 1000 = 139. 

Example. — From a water power it is desired to run a 
circuit 14 miles to supply a compact district with 500 
K.W. in incandescent lamps, the current being distributed 
in the town on the three-wire, low-tension system from a 
transformer substation. Station pressure is 130 volts at 
full load, lamps 120 volts. The maximum potential is 
about 10,000 volts. For the substation 40 K.W. trans- 
formers are used in seven pairs, ratio 10 to 1, primaries 
connected two in series, and their secondaries connected 
for the three-wire system. The frequency is 60. The 
line is No. 00 B. & S. If we may assume a power 
factor of 0.99, and an inductance factor of 0.14, what will 
be the values corresponding to those obtained in previous 
example ? 

Solution. — First reducing secondary circuits to main 
line conditions, we obtain 9100 volts and 59.7 amperes. 
By use of the table we obtain the line resistance 0.412 
ohm per mile, inductive resistance at 12 inches apart 
0.534 ohm per mile ; and adding 15% this becomes 0.614 
ohm per mile. 



ALTERNATING CURRENT DISTRIBUTION. 2 J I 

On secondaries, 

0.99 X 9100 = 9000 active volts. 
Also 0.14 X 9100 = 1270 inductive volts. 

/== 59.7 amperes. 

Reducing transformers, 

Ohmic drop i°/ of 91 00 = 91 volts. 
Inductive drop 6°/ of 9100 = 546 volts. 
Core loss, say 3% of 59.7 = 1.78 amperes. 

Main line, 

Ohmic drop 0.412 X 28 X (59.7 + 1.78) = 7 10 volts. 
Inductive drop 

0.614 X 28 X (59.7 + 1.78) = 1060 volts. 
Total active volts 

E a = 9000 + 91 + 710 = 9801. 
Total inductive drop 

E s = 1270 + 546 + 1060 = 2876 volts. 

Therefore the voltage to be supplied at the secondaries 
of the step up transformers is 

2 2 

9801 +2876 = 10,200 volts. 

The current supplied by step up transformers is 

/ = 59.7 + 1.78 = 61.5 amperes. 

Ohmic drop i°/ of 10,200 = 102 volts. 

Inductive drop 6°f of 10,200 = 612 volts. 

Core loss $°/o of 61.5 = 1.85 amperes. 

Therefore the active voltage supplied by the gen* 
erator is E a — 9801 + 102 = 9903 volts, 

And E s = 2876 + 612 = 3488 volts. 



272 ELECTRICAL AND MAGNETIC CALCULATIONS. 



Therefore E { =z^/ggo^ 2 + 3488 s = 10,500 volts. 
And / = 61.5 + 1.85 = 63.4 amperes. 

Hence the generator delivers 10,500 X 63.4 = 665,700 
volt-amperes, and 10,500 X 59.7 = 626.8 K.W. 

The power delivered at the transformer station is 
\%% X 130 =541 K.W., and the efficiency of the plant 
from generator to step-down secondaries is 541-5-627 
= 86%. 

Sixteen of the 40 K.W. transformers connected two in 
series to the generator, and two in series to the line are 
used as step-up transformers, receiving 1050 volts and 
delivering 10,500 volts. 

The generator capacity is 665 K.W., and allowing 92% 
efficiency, there will be required in water power 665 -r- 
0.92 = 685 K.W. Total efficiency is then 73%. 

64. Formulae and Tables for Alternating Current Wir- 
ing. — For most practical cases the following formulae 
based on Ohm's law will be very convenient and suffi- 
ciently accurate for the calculation of transmission cir- 
cuits. The constants which are introduced in the formulae 
have the values under the different conditions given in 
the proper tables ; these are taken from the publications 
of the General Electric Company. 

Symbols. — / = total line current. 

E = E.M.F. at customer's end of circuit. 
W— Watts delivered to the customer. 
oJ = Per cent of W loss in line. 
D = Distance of transmission. 



ALTERNATING CURRENT DISTRIBUTION 273 



K= Factor depending on the power factor and 

system used ; given in table ; equals 2160 

for single phase, 100% power factor. 
T= 1 for continuous currents, and depends on 

power factor and nature of system used ; 

given in table. 
Jlf= Factor depending on frequency, size of wire 

and power factor ; equals 1 for continuous 

currents ; given in table. 
A = Factor based on 0.00000302 lb. as weight 

of 1 mil-foot. 
Formula. — _ W 

I=TX -E- 

WD 

Area, circular mills, d 2 = K X 



^X % 



£ 2 x % 



Volts line loss = M X 

100 

A X WxKxD 2 



Pounds copper = 



£ 2 x % x io ( 



For continuous currents, 

T= 1, M— 1, A = 6.04, K = 2160. 

11. Table of Wiring Constants. 



(138) 

(!39) 
(140) 

(i4i) 





Values of T. 


Values of K. 





Per Cent 


Per Cent 


System. 


Power Factor. 


Power Factor. 


73 

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1-1 
< 

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95 


90 


85 


80 


100 


95 


90 


85 


80 
338o 


Single-phase 


I.052 


I. II 


1. 17 


I.25 


2160 


2400 


2660 


3000 


6.04 


Two-phase, 






















4 -wire . . 


.526 


•555 


.588 


.625 


1080 


1200 


I330 


1500 


1690 


12.08 


Three-phase, 






















3 -wire . . 


.607 


.642 


.679 


•725 


1080 


1200 


l 33° 


1500 


1690 


906 



274 ELECTRICAL AND MAGNETIC CALCULATIONS. 

To find the value of K for any other power factor, use 
the following : — 

For single-phase, K = — , , N2 • (142) 

(Power factor) 2 v * ' 

-r, 1 • 2160 , N 

For two-phase, 4-wire, K = —7=- - ^- (143) 

r 2 (Power factor) 2 v ^ OJ 

For three-phase, 3-wire, K = — - - ^= • (1 44) 

^ ' ° 2 (Power factor) 2 v w 

Example. — Suppose it is required to find K for 3- 

phase lines for a power factor of 75, say a circuit largely 

of motors partly loaded, — 

2160 
a»= —. — V2 = 1920. 
2 (-7S) 2 

Calculation of Circuits by the Formula. — Direct 
Current, Two -Wire Circuits. 

Example. — Determine the size of wire and the loss of 
E.M.F. to supply 1000 lamps, no volts, at a distance of 
800 feet from the generator, loss 10%. 

Solution. — 
__ K X W XD _ 2160 X (1000 x|.Xiio)x 800 
"" £ 2 X % no 2 X 10 

= 785,450 cir. mils. 

TTU . , MxEx°/ 1X110X10 

Volts lost = '— = = 11 volts. 

100 100 



Weight of copper = — — — -, ^ — 

8 rr £ 2 X%X io 6 



6.04 X (1000 XiXno)X2i6o X800 „ 

= — V 2 J — m =3795 lbs. 



no X10X10' 



ALTERNATING CURRENT DISTRIBUTION. 2?$ 



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276 ELECTRICAL AND MAGNETIC CALCULATIONS. 

Direct Current Three-Wire Circuits. 

Example. — If 500 lamps, no volts, one-half ampere, 
are fed over a line 1000 feet long at a loss of 5%, three- 
wire system, what size and weight of wire will be 
necessary? 

Solution. — 

„_ KxWxD _ 2160 X (500 X \ X no) X 500 

~~ £* X % = ~ Sx S 

= 122,750 cir. mils. 

_ MxE x % 1 X 220 x s 

Drop = '- = 2 = 1 ! volts. 



Weight = 



100 too 

A XWXK XD 2 



& X % X 10 6 
_ 6.04 X (500 X i X no) X 2160 X 500 2 

220 2 x s x io 6 
= 370 lbs. 

If the neutral is one-half the size of the outside wires 
the area would be J of 122,750 = 61,370 cir. mils, and 
its weight would be ^ of ^|^- = 93 lbs., making the total 
weight 370 + 93 = 463 lbs. If the neutral is to be the 
same size as the outside conductors it will add 185 lbs., 
making 370+ 185 =555 lbs. If this is the secondary 
of a transformer system the E.M.F. of each of the two 
transformers arranged in series to supply the 3-wire line 
must be 115.5 volts ; or if from dynamos direct, they must 
each deliver 115.5 volts. 

Alternating Current, Single-Phase, 60 Cycles. 

Example. — Find the size of wire and line drop for the 
single-phase, 2-wire primary feeders for 1200, 3.5 watt 



ALTERNATING CURRENT DISTRIBUTION, 277 

no volt lamps, transformers 20 to 1. From generator to 
transformers 2400- feet. Transformer drop 3%, and in 
secondary wiring 2%, primary line loss 5% of delivered 
watts; transformer efficiency 97%. 

Solution. — 

Power at lamps = 1200 x (3.5 X 16) = 67,200 watts. 

-r-» r . 67,200 

Power at transformer primary = — = 70,690 watts. 

E.M.F. at transformer primary 

= (110 + 2% of no) X 1.03 X 20 = 2311 volts. 
Cross section, 

_iTx W X D _ 2400 x 70,690 x 2400 

"" & X % ~~~ 2311 2 x s 

= 15,250 cir. mils. 

This assumes power factor =95, making K = 2400. 

The nearest size is No. 8 = 16,500 cir. mils. 

Loss °/ on primary mains, using No. 8, is 

. Ky.Wy.D_ 2400 x 70,690 x 2400 _ 

'° " & X d 2 " 2311 2 X 16,500 = 4 

, MyEyo/o 1.02x2311x4.62 

Line drop = — = 

100 100 

= 109 volts. 

Generator E.M.F. = 2311 + 109 = 2420 volts. 

Current in primary circuit is 

_ T y W 1.052 x 70,830 
I = — — = — — = 32.2 amperes. 

Single-Phase, 125 Cycles. 

Example. — Solve the above problem for a 125 cycle 
circuit. 



278 ELECTRICAL AND MAGNETIC CALCULATIONS. 



Solution. — Since M is the only factor which changes 
with frequency, the solutions and results will be the same 
for all frequencies except the line drop, and therefore the 
machine volts. 



T . , M X E X °) i-°9 X 2311 X 4-62 

Line drop = — = z 

100 100 

= 116.37 volts. 
Generator E.M.F. = 231 1 -f 116.37 = 2 4 2 ^. 



Two-Phase, Four-Wire, 60 Cycles. 

Example. — Calculate the line to transmit 3000 H.P. 
4 miles to step-down transformer secondaries. Generator 
supplies such E.M.F. as will give 6000 volts at step-down 
primaries, and the line loss is to be about 10% of delivered 
power. Transformer efficiency will be 97% and load of 
such character as to make the power factor about 85%. 



k 



11 



n Mii w 



oamrn 



uaaa 

M 



Fig. 41. 



Solution. — 

Power at secondaries = 3000 H.P. = 2238 K.W. 

2238 



Power at primaries = 
Line loss =10%; 



•97 



= 2307.2 K.W. 



ALTERNATING CURRENT DISTRIBUTION 279 

Hence 

_Kv.Wy.D_ 1500 x 2,307,200 x (4 x 5280) 
" E 2 X % 6W x io~ 

= 203,040 cir. mils. 
Take 4 No. 3 wires in parallel = 52,600 X 4 = 210,400 
cir. mils. 

Loss on mains using 4 No. 3 wires in parallel making 
16 wires in all is 

Ky.Wy.D_ 1500 x 2,307,200(4 x 5280) 
'° ~~ j£ 2 X d 2 : " 6000 2 X 210,400 

= 9.65%. 
Power lost in transmission = 3000 x .0965 = 2^9.5 H.P. 

___._. M X E X °/ 1. 18 X 6000 X 9.65 

E.M.F. loss = '— = — - 

100 100 

= 683.2 volts. 
Generator E.M.F. = 6683 + volts. 

r TW .588 X 2,307,200 

Line current I = —=- = — , = 226 amperes. 

E 6000 

Add for transformer core loss ii°/ making 230 amperes. 

A x WyK yD> 

Copper = — — -. = — 

12.08 X 2,307,200 X 1 coo X (4 X C280) 2 r „ 

= ==q — r e — — = S3 ' 67S lbs - 

6ooo 2 X 9.65 X io 6 
Two-Phase, Three- W ire Distribution* 

This case may be conveniently worked out by finding 
first the weight of the copper and size of wire for the 
single-phase, or two-phase, four-wire system ; then when 
the E.M.F. between the common middle wire and either 

* See Crocker's "Electric Lighting," Vol. II., page 228, for the derivation of 
the constants used in this case. 



280 ELECTRICAL AND MAGNETIC CALCULATIONS. 

outside wire is the same as in other cases, that is, when 
the minimum E.M.F. is the same, the weight of copper in the 
two-phase, three-wire system will be 7J°f °f that found for 
the single-phase, or two-phase, four-wire system. Each out- 
side wire will have 42.7 °J of the area of each in the single- 
phase, and 8^.4°f of each in the two-phase, four-wire system. 
The middle wire will have 6o.4°j and 120.8 °f respectively. 
In the last example the weight of copper required was 
53,675 lbs. 

In this system it must be 

73% of 53,675 = 39,182.75 lbs. 

The cross section for each outer conductor will be 
d 2 = 85.4% of 203,040 = 173,396 cir. mils. 
= No. 000 wire, nearest. 
The common wire will be 

d 2 = 120.8% of 203,040 = 245,272 cir. mils. 
= 3 No. 1 wires in parallel. 

When the E.M.F. between the two outside wires is the 
same as the other systems, that is, when the maximum 
E.M.F. is the same, the rules for the weight of the copper 
and size of conductors are as follows : 

The weight of copper will be 145.7 °/ of that found as 
directed above. Each outside wire will have <5 ) J'.5% of the 
cross section of each in the single-phase and I7i°f °f ec ^ cn in 
the two-phase, four-wire system. The common middle wire 
requires. 120. 4°j and 240.8 °f respectively. 

Assume 6000 volts pressure between the outer wires in 
the two-phase, three-wire system ; the weight of copper 



AL TERNA TING CURRENT D IS TRIE UTION. 2 8 I 

required under the conditions given in last example will 
be 145.7% of 53,675 =78,204 lbs. The outer wires will 
have d 2 = 171% of 203,040 = 347,198 cir. mils. This 
corresponds most nearly to 2 No. 000 wires in parallel. 
The middle wire will have 240.8% of 203,040 = 488,920 
cir. mils. Take 3 No. 000 wires in parallel. 



Three-Phase, Three-Wire Transmission, 60 Cycles. 

Example. — Solve the last example for three-phase, 
three- wire circuits. 

T 



I 



So. 

D 




Fig. 42. 



Solution. — Power delivered at secondaries, 2238 K.W, 
At primaries, 2307.2 K.W. 

i^__K xW x D _ 1500 x 2,307,200 x (4 x 5280) 

I? X °j 6000 2 X 10 

= 203,040 cir. mils. 

Take 4 No. 3 wires in parallel for each conductor, mak- 
ing 12 in all, giving d 2 = 52,600 X 4 = 210,400 cir. mils. 

, _K XW X D __ 1500 X 2,307,200 X (4 X 5280) 



&Xd 2 
= 9- 6 5- 



6000 X 210,400 
Power lost on line = 3000 X .0965 = 289.5 H.P. 



282 ELECTRICAL AND MAGNETIC CALCULATIONS. 
E.M.F. lost = M * E *°1° = '-18X6000X9.65 

IOO IOO 

= 683.2 volts. 

Generator E.M.F. = 6683 + volts. 

TW .679 X 2,307,200 

1= —=- = — — — , ° ' = 261 amperes. 

E 6000 

Add \\°j f° r core loss, making /= 265 amperes. 

. , A X WxKX U 2 

Copper required = — — -. -„ — 

v * M £? X % X io 6 

9.06 X 2,307,200 X 1500 X (4 X 5280) 2 

^ 2 ~7 6 

6000 X 9.65 X 10 

= 40,259 lbs. 
Three-Phase, Four-Wire Distribution, 60 Cycles. 

Example. — The secondaries from transformers to 
center of distribution are 500 feet long. The load con- 
sists of 5 20 H.P. induction motors, 200 volts, and 1000 
16 c.p., 115 volt lamps, \ ampere each. Allow about 
1 1 volts drop on lighting side from transformers to center, 
and 5 volts from center to motors. The motor efficiency 
may be assumed at 80 per cent and power factor 80 
per cent. The E.M.F. at distributing center between the 
main wires is 205 volts. Determine the wiring as in pre- 
vious examples. 

Solution. — For the motors, 

TX W .725 X 4 X 20 X 746 

/ =: — — - = J— ^ — l — = 270.5 amperes. 

1 E x eff . 200 X .80 ' ° 

For the lamps, 

T X W .607 X (ioooX^X 115) 
I 2 = — = 1 i l - ^ = i74.S amperes. 

±L 200 

/= I x + I 2 = 445 amperes. 



ALTERNATING CURRENT DISTRIBUTION. 283 



nnnM 



200 




n 

in 2 



NEUTRAL 



-o- -o- 



n 



-o -o 



Fig. 43. 



For the motors, 



w _ A£ = 270-5x205 = 76>5 KWi 



725 



For the lamps, 



$B = ^74-5X205 = KW< 

y .607 

^=76.5 + 58.9= I3S-4 K.W. 
The drop will be the equivalent of about 6%. Hence 
W X D 1200 x 58,900 + 1690 x 76,500 



d 2 = K 



135,400 X 500 



i35>4-oo 
= 396,500 cir. mils. 



205 x 6 

Take 3 No. 00 wires in parallel = 3 X I 33>°79 = 
399,237 cir. mils. 

_ 1200 x 58,900 + 1690x76,500 135^400 X 500 

jo — X o 



= 5.9%. 



135^400 



2 °5 ^399,237 



284 ELECTRICAL AND MAGNETIC CALCULATIONS. 

,r 1 1 ,^ X °/o 1.62 X 205 X K.g 

Volts lost = M - = 5 ^_Z = Ig> 6. 

100 100 

E.M.F. at transformer secondaries = 205 + 19.6 = 
224.6 volts. 

Drop to center measured between each main and the 
neutral 

= 19.6 x H# == I1 - 2 7 volts. 

Cross section of neutral conductor need be only 

d 2 = }1±5_ x (3 X 133,079) = i55>5°° cir - mils - 
445 

Use No. 000 = 167,800 cir. mils. 



Monocyclic, 60 Cycles. 

Example. — The load consists of 4 20 H.P. no volt 
induction motors, 85% efficiency, and 2000 half-ampere, 
104 volt incandescent lamps. Distance of transformers 
from generator, half a mile. Motors are 100 feet from 
transformers, line loss 4%. Transformer efficiency 97%. 
Primary line loss 5%. No load E.M.F. of generator 1040 
volts. Separate transformers are used for lights and 
motors. Determine lines as before. 

Solution. — At motors 

4 X 20 X 746 

W = - — - — = 70,212 watts. 

0.85 

For 80% power factor 

t -WXjD ' 70,212X100 . ., 

d 2 = K = 3380 X ' ! , =490,000 cir mils. 

£ 2 X°/o no 2 X4 



ALTERNATING CURRENT DISTRIBUTION. 285 



Two No. 0000, or 4 No. o wires may be used for each 

conductor from transformers to motors. The latter will 

have the advantage of giving the smaller voltage drop for 

the same loss in watts. The exact proportion of drop 

.„ , 1.28 . , 

will be — — = 69.2 %. 

j 

Using No. o wires, 

, ^Wy.D 70,212x100 , 

# = K W*!> = ^° X 4 X ,05,593 X no* = 4 - 6 ^' 

E.M.F. drop on secondaries to motors 

.-jExtfo 1.49 Xi 10X4.6 . 

— M '— = — — — = 7.50 volts. 



100 



100 



UJU 



cm 



iop: 



2640 ft 









SLMJ 



nro 



Fig. 44. 

Primary E.M.F. at motor transformers 

= (no + 7.50) X 9 X 1.03 = 1089 volts.* 

* See p. 163 for ratio of transformation. 



286 ELECTRICAL AND MAGNETIC CALCULATIONS. 

Hence E.M.F. at secondaries of lighting transformers, 

1089 , 

= -T- 1.03 = 105.7 volts. 

Also at motor transformers 

___ 70,212 X 1.046 

W ^ ~Y7^ 7^V~ = 7S ? 7i3 watts. 

(1.00 — .03) 

At lighting transformers 

2000 X^X 105.7 • 

W 2 = — = 108,970 watts. 

0.97 '*' 

W ' = 108,970 + 75,713 = 184,683 watts. 
Total power factor 

108.9 X .95 + 75-7 X .80 
= i8^" - 88 %' 

Hence ^ = 2160 -s- .88 =2424. 

From primary circuit 

. ^WxD 184,683 X 2640 

d 2 = K = 2424 X ° 9 — 

E 2 X % 1089 2 X 5 

= 200,000 cir. mils. 

Take 2 No. o wires in parallel, making 

, 2424 X 184,683 X 2640 
oj = 2 = 4.0. 

1089 X (2 X 105,592) 
Primary drop 

_ MxE X °J C __ 108.9 X 1.30+75.7 Xi.49 1089 X 4-8 
100 184.7 IO ° 

= 72 volts. 

Generator E.M.F. = 1089 + 72 = 1161 volts. 

^ T I l6l — I040 , , 

Compounding = = 11.6%. 

1040 ' 

_ TxW 1.099 X 184,683 

I = ^ — = — s = 186 amperes. 

E 1089 r 



ALTERiYATIXG CUR RE XT DISTRIBUTIOX. 287 



For teazer wire 



211,184 X 



75-7 



= 141,300 cir. mils. 



108.9 
Use 3 No. 3 wires in parallel. 

Monocyclic, Three -Wire Secondary for Lights and 
Motors, 60 Cycles. 

Example. — From generator to transformer the dis- 
tance is 2000 feet, and 200 feet from transformer to 
motors on each circuit, and 200 feet also to center of 
lights. Drop on secondary mains about 10 volts, on 



i, . | 2000 FT. 
On !f r -i- 

Dl_J J_L_J 



jujJZuLuUju. 



: nnrrnnrLpm 

/g i |<+88-> | I ■ « 229>r— > 



220 




220 



M 






Fig. 45. 

primary mains 3.5%, in transformers 3.5% ; energy loss 
in transformers 3%. The load consists of 2000 no volt 
lamps and 3 20 H.P. induction motors, 86% efficiency. 

Solution. — For lamps 

W x = 2000 X \ X no = 110,000 watts. 



, volts loss X IOO „ . 

%= — ^r^ — = 2 - 8 % 



288 ELECTRICAL AND MAGNETIC CALCULATIONS. 

For 10 volts secondary loss 

s loss X 

Therefore 

79 T . W X D 110,000X200 
</* = X = 2400 x — =rs 

-S 2 X % 220 X 2.8. 

= 389,610 cir. mils. 

Take 3 No. 00 = 3 x 133,079 cir. mils. 

, T ^W X D 110,000 X 200 

% = K ,^ m = 24 °° X ^~=* = 2 ' 7 ' 

d z x E z 399> 2 37 X 22 ° 

_ __^ X % 220 X 2.7 

Drop = Jf '— = 1.34 X = 9 volts. 

100 100 

Taking neutral = \ of outside = one No. 00. 

E.M.F. at transformer secondary =229 volts. 

In secondary lighting mains 

T rnW 110,000 

7i= ^-et = 1-052 x = 526 amperes. 

Amount of copper in 1400 feet of No. 00 = 666 lbs. 
For motors 

___ 20 x 746 X 3 , 

2 = 086 = 52, ° 4 

= 17,348 watts on each circuit. 

In order to make the drop on motor circuits the same 
as lighting mains for 80% power factor, the °f drop 
must be taken at about 3.5%. Hence 

79 o I 7>348 X 200 . .. 

d 2 = 3380 X \ = 70,210 cir. mils. 

220 X 3.5 
Take one No. 2 = 66,373 cu ~- m ils. 

, KxWxD 17,348x200 , . 

E 2 X d 2 220 X 66,373 



ALTERNATING CURRENT DISTRIBUTION 289 

220 x ^ 6 

E.M.F. drop = 1.26 X — = 10 volts. 

100 

This is a little greater than the drop on lighting mains, 
and may be reduced by taking the next larger size of 
wire ; but for the operation of motors this slight differ- 
ence is immaterial. 

^W 52,046 
I 2 = T— = 1.25 X = 296 amperes. 

For the total load 

W= 110,000 + 52,046 = 162,046 watts. 

Total power absorbed by transformer 

162,046 

= = 167, 058 watts. 

0.97 " * 

Primary E.M.F. = 229 X 1.035 X 9 = 2134 volts. 

For primary feeders 

110,000X2400 + 52,046x3380 167,058x2000 

167,058 2i 3 4 2 X3.5 

= 55,284 cir. mils. 
Take No. 3 = 52,633 cir. mils. 

WxD 2634X167,058x2000 

% =K j?2 „ ^ = =2— = 3-67- 

E 2 xd 2 2134 X 52,633 

_ , 110,000X95+52,046x80 . , 

Power factor = ^ V— = 87%. 

167,058 ' 



.. , MxEx°i 1.175 X2134X3J 
Primary line drop = '- X — — — — — 



_ 6 7 
100 100 

= 91 volts. 
Generator E.M.F. = 2225 volts. 

-r. • 167,058 

Primary current = 1.14 X — - — — = 80.2 amperes. 

2134 

For teazer d 2 = 55,284 X ^- 1 = 26,150 = No. 6 wire. 

009 110,000 ° 



29O ELECTRICAL AND MAGNETIC CALCULATIONS. 

65. Original Problems. — 1. Neglecting the effect of 
induction calculate the size of the primary mains and also 
the secondary leads of a large transformer fully loaded 
constantly, whose efficiency may be taken at 98%. The 
secondaries run 200 feet to feed 1000 lamps at 5% loss. 
The primaries run \ mile at 5 % loss. The ratio of trans- 
formation is 20. The lamps are no volts. Determine 
voltages and capacities. 

Secondaries a No. 0000 and a No. 000 wire. 
Primaries No. 10 wire. 
Voltage at secondary terminals 116 volts. 
Voltage at primary terminals 2340 volts. 
Secondary current 500 amps., primary 2.53 

amps. 
Machine volts 2463, primary drop 123 volts. 
Machine capacity 65 K.W. 

Transformer capacity 60 K.W. ; lights 55 K.W. 
2. A machine of 60 K.W. capacity is to supply two 
fully loaded transformers at an average of f mile. The 
secondaries are respectively 200 feet for § total load, and 
100 feet for \ total load. Losses are 5% on secondaries, 
5% on primaries. Transformer efficiencies, 97% for the 
larger, 96% for the smaller. Determine capacity of trans- 
formers and lights ; also the sizes of primary and second- 
ary wires. The ratio is 10. 

Load at transformers 57 K.W. 

Secondary, larger, 37 K.W., smaller, 18 K.W. 

Load lamps 52 K.W. Primary wire No. 2. 

f a No. o and a No. 00 in 
Secondary wire -\ parallel on larger. No. 2 

wire on smaller. 



ALTERNATING CURRENT DISTRIBUTION. 29 1 

3. A load of 1000 incandescent no volt lamps is put 
on the transformer secondaries ; the transformers are of 
nearly one size and are so loaded that about S c / will 
represent the inductance loss and 1 °f the resistance loss. 
The core loss will be 5%. Also 4% line loss is allowed 
in the secondary wiring. The primary wire is 5 miles long, 
18 inches apart, and is No. 00, carrying the current at 
125 cycles. Determine the generator volts, transformer 
pressure, both primary and secondary ; also the ohmic 
and inductive drops. Ratio 10. 

Voltage 1 100. 
Current 50 amperes. 
Secondary ohmic drop 44 volts. 
Secondary inductive drop 3%, 33 volts. 
Transformer ohmic drop 1%, n volts. 
Transformer inductive drop 8°f 0y 88 volts. 
Core loss 5% of 50 amperes = 2.5 amperes. 

/(total) = 52.5. 
Line ohmic drop = (0.4 1 2 X 10) X 52.5 = 2 16.3 volts. 
Line inductive drop = (1.23 X 10) x 52.5 

= 6 45-75 + 1 5% = 742.6 volts. 
Total active volts = 1100 + 44 + n + 216.3 

= I 37 1 -3- 
Total inductive volts = 33+88 + 742.6 = 863.6. 

Total impressed volts = ^li^i.^ + 863. 6 2 

= 1626 volts. 
Generator apparent watts = 84365. 

Secondary impressed volts = V1144 2 + 33 2 

= 1 144.5 v °lts. 

Primary impressed volts = vnj^ 2 + 121 2 

= 1 1 61. 3 volts. 



292 ELECTRICAL AND MAGNETIC CALCULATIONS. 

4. A long-distance plant generates about 1200 volts, 
steps up to about 12,000, carries the power over 16 miles 
of No. 1 wire to sub-station, where it is transformed down 
to about 1000 volts, then carried over wires and distances 
equivalent to 2 miles of No. 000 wire, then transformed 
for consumers to 100 volts. The frequency is 60 and the 
ratio of transformation is 10. Assume a power factor in 
the secondaries and lamps of 98.5, thus giving an induc- 
tance factor of 17. Allow 1% ohmic loss in all trans- 
formers and 6°/ average inductive drop. These per cents 
are taken on secondary voltages. Core losses are figured 
at 3%. If the energy delivered is equivalent to 9000 
lamps at 50 watts each, make the calculation of pressures 
and generator energy. 

In terms of high voltage line the secondary E.M.F. is 
10,000 volts and the current is 50 amperes. 

Active secondary E.M.F. = 98.5% of 10,000 

= 9850 volts. 
Inductive secondary E.M.F. = 17% of 10,000 

= 1700 volts. 
Step-down transformers, 

Ohmic drop \°j of 10,000 =100 volts. 
Ind. drop 6% of 10,000 = 600 volts. 
Core loss 3% of 50 amperes = 1.5 amperes. 
Total current in primaries = 50 + 1.5 = 51.5 amp. 
Primary impressed E.M.F. 

£ { = V(985o +ioo) 2 + (i7oo + 6oo) 2 =io,2i2 
volts. 
Line between step-down transformers, 

Ohmic drop = (0.324 X 4) X 51.5 = 67 volts. 
Ind. drop = (0.607 X 4) X 51.5 = 125 volts. 



ALTERNATING CURRENT DISTRIBUTION 293 

Total voltage at secondaries, step-downs, 

E L = Vio,oi7 2 + 2425 s = 10,306 volts. 
First step-down transformers, 

Ohmic drop = i c f c of 10,306 =103 volts. 

Inductive drop = 6 c / of 10,306 = 618 volts. 
Volts impressed on primaries, step-downs, 

E t = v IO ,i2o 2 + 3043 = 10,572 volts. 
Current in long-distance line, 

/= S 1 ^ + 3 c /o of 51.5 = 53 amperes. 
Long-distance line, 

Ohmic drop = (0.65 X 32) x 53 = 1 100 volts. 

Ind. drop = (0.65 X 32) X 53 = 1100 volts. 
Voltage at secondaries, step-ups, 

11,220 +4143 = 11,998 volts. 
Step-up transformers, 

Ohmic drop = i°/ of 11,998 =120 volts. 
Inductive drop, 6% of 11,998 = 720 volts. 
Primary impressed pressure, 

E t = Mi,34o 2 + 4863 s = 12,339 volts. 
Primary current from machine, 

I= = 53 + 3 c /o oi S3 = 54-6 amperes. 
Apparent power = 674 K.W. Lag 23 . Cos 23 

= 0.918, total power factor. 
True power = 674 x 0.918 = 618.7 K.W. 
Total transmission loss 

= 618.7 — (9.000 X 50) = 163.7 K.W. 
Total per cent transmission loss 

= 163.7 -J- 618.7 = 26.46%. 

. Solve problem 5, page 80, by the formulae of 64. 



294 ELECTRICAL AND MAGNETIC CALCULATIONS, 

6. Determine the wiring for the following single- 
phase, 125 cycle circuit. From generators to trans- 
formers 2000 feet, loss about 5% of delivered power. 
Transformer drop 3%, energy loss 3%. Drop in sec- 
ondary wiring 2 volts. Ratio of transformation 10 to 1. 
Load consists of 1000 16 c. p. 3.6 watt, 104 volt lamps. 

Wire No. 3 B. & S. 
Primary drop = 68.4 volts. 
Generator E.M.F. = 1160. 
Primary current = 58.3 amperes. 

7. The distance from generator to transformer is 
2500 feet, loss 5%. Ratio of transformation 20 to 1 ; 
efficiency of transformer 97-3-%, voltage drop 2%. Sec- 
ondary loss 2 volts. Load 750, 52 volt 60 watt lamps. 
Determine the wiring, etc., for a single-phase, 60 cycle 
system. 

Wire No. 3 B. & S. 
Primary drop = 54 volts. 
Generator E.M.F. = 1155.4 volts. 
Primary current = 44.8. 

8. Calculate the transmission circuit for a two-phase, 
four-wire, 60 cycle system to deliver 2500 H.P. 5 miles 
at 6000 volts, line loss 7^%. Load consists of induction 
motors. Efficiency of transformers 97^%. 

Wire 3 No. o B. & S. 
Line loss = 574 volts. 
Generator E.M.F. = 6574 volts. 
I = 199 amperes. 
Core loss = 3 amperes. 



ALTERNATING CURRENT DISTRIBUTION 295 

9. In a two-phase, four-wire system, 125 cycles, 5000 
H.P. is transmitted si miles at IO % loss - The E.M.F. 
at transformers is 5000 volts, and the load has a power 
factor of about 85%. Determine the size of wire and 
generator E.M.F. 

Wire 4 No. o B. & S. 

Line loss = 9.79%. 

E.M.F. drop = 1042.6 volts. 

Generator E.M.F. = 6042.6 volts. 

10. Solve 8, assuming a three-phase, 60 cycle trans- 
mission circuit. 

Wire 3 No. o B. & S. 
Line drop = 574 volts. 
Generator E.M.F. = 6574 volts. 
/= 237.5 amperes. 

11. Solve 9 again, assuming the same conditions for a 

three-phase circuit, three-wire, but having a frequency of 

60 cycles instead of 125. 

Line 4 No. o wires. 

Drop = 646 volts. 

Generator = 5646 volts. 

/ = 514 amperes. 

12. A three-phase, 60 cycle, four-wire system of distri- 
bution is to be constructed to supply a load of 750 115 
volt lamps, and 4, 16 H.P., 200 volt induction motors. 
From transformers to center of distribution the distance 
is 600 feet. There are to be 15 volts drop on lighting 
circuits between transformers and center, and five volts 
from center to motors. The motor efficiency will be about 
85%, and E.M.F. at the center about 205 volts. 



296 ELECTRICAL AND MAGNETIC CALCULATIONS 

Lamp circuits 2 No. o wires. 

Drop = 26.4 between mains, or 15.2 

between mains and neutral. 
Neutral wire = No. 1 B. & S. 
For motors, current =191 amperes. 
For lamps, current =191. 
/= 392 amperes. 

13. At a distance of 3000 feet from the generator a 
transformer is stationed to supply 1500 half ampere, 104 
volt lamps and one 25 H.P. no volt induction motor, 
efficiency 85%, system monocyclic. From transformer to 
motor the distance is 100 feet, loss 2^%. Efficiency of 
transformers 97%. Primary loss 4%. Determine the 
wiring, etc, as before, frequency 60 cycles. 

Motor circuit 2 No. o wires. 
Drop in motor circuit = 4 volts. 
Lighting circuits No. 000 B. & S. 
Drop primary circuits = 68.5 volts. 
Generator E.M.F. = 1 144.5 volts. 
1= 106 amperes. 
Teazer No. 4 B. & S. 

14. A load of four 10 H.P. induction motors at a dis- 
tance of 200 feet from transformers, and 1000, no volt 
lamps, 16 c. p., at a distance of 150 feet from trans- 
formers, fed by a three-wire secondary system, mono- 
cyclic, 60 cycles, are supplied with power by a generator 
1000 feet from transformers. If the primary drop is to 
be about 3% and the drop in secondary circuits about 10 
volts, determine the wiring, etc., assuming transformer 
drop 4% and energy loss 3%. 



ALTERNATING CURRENT DISTRIBUTION 2g) 

For motor circuits, wire is No. 5 B. & S. 
Volts loss = 8.25. E.M.F. = 220 volts. 
For lights, wire = No. 000 B. & S. 
Drop = 8 volts. E.M.F. = 220 volts. 
Neutral = No. 2 B. & S. 
Primary feeder = No. 7 B. & S. Teazer 

= No. 8. 
Primary current = 48.8 amperes. 
Primary drop = 58.5 volts. 
Generator E.M.F. = 2192.5 volts. 



298 ELECTRICAL AND MAGNETIC CALCULATIONS. 





13. Magnetic 


Properties (a). 


Annealed Charcoal Iron. 


B Gausses. 


H Gilberts per 
cm. Length. 


/x Permeability. 


-j- Amp .-Turns 
per cm. Length. 


3,000 


i-34 


2,238 


I.07 


4,000 
5,000 


i-55 
1.76 


2,580 
2,841 


I.24 
1. 41 


6,000 


2.03 


2 >955 


I.62 


7,000 


2.41 


2,904 


i-93 


8,000 
9,000 


2.96 
3-56 


2,703 
2,528 


2-37 
2.85 


10.000 
1 1 ,000 
12,000 


4-37 
5.60 
7.60 


2,288 . 

1.963 

1.579 


3-5° 
448 
6.08 


13,000 


11.30 


1,151 


9.04 


14,000 
15,000 


16.90 
28.60 


839 
524 


13.50 
22.80 


16,000 


48.80 


321 


39.00 


17,000 


100.00 


170 


80.00 


18,000 


186.00 


96 


148.00 


19,000 


350.00 


54 


280.00 


20,000 


650.00 


3 1 


520.00 



Magnetic Properties (b). 





Soft Gra^ 


Cast Iron. 




B Gausses. 


H Gilberts per 
per cm. Length. 


/a Permeability. 


—j- Amp .-Turns 
per cm. Length. 


3,000 


I.50 


2,000 


I.20 


4,000 


5-50 


727 


4.40 


5,000 


IO.70 


467 


8.60 


6,000 


23.20 


258 


18.50 


7,000 


43.20 


162 


34.60 


8,000 


78.50 


102 


62.80 


9,000 


123.70 


72 


98.90 


10,000 


188.50 


53 


150.80 


1 1 ,000 


288.OO 


39 


230.40 



ELEMENTS OF DYNAMO DESIGN. 



299 



CO 

53 
O 
i-h 

H 
O 

U 

1— 1 

u 

o 
z 



3 
H 





< 


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o 6 


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66666666666666 6 6 6 6 6 6 6 6 6 


H 

72 


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H 


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low rON«Ors n^tO\ON-tuiN *>">vO cooo •-" mrJ-^NO 
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n ^vo 00 O fO"^NO N Tfr^O N tnoo O cono O conO O 
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60000000000000' dddddo' dciw 


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rOTj-Tj-Tj-Tf'^-'^-i-oi-ovnvovotn lovq O \D ^O 'O \0 ^ *0 r>» 

oodo'o'o'oddddddoo'ddddoddd 


w 
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U 




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C3 
< 


6 
z 


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onoo 000000000000000000 t^t^.t^r^t>»t^r^r^t^ t^vo vo 


o 6 


(A 

U 


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mOOOOOOOOOOOOOOO 0000000 


H 


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H 
O 

u 


O O f^H nh rj-ro Tj-00 CONO NO tOOO i-h rJ-ONt^N i^m i-h 
O ONO hi O O rf" rj- VOCOH tTtJ-i-i O N t^O N^j-N 1 ^ 1 ^ 
O ^J- cooo O CO "-H TfM hh NrfO com COCO r>. t^ O Tt- O r^ 
O ^ NO O ^ "<f ^ •-; hH ro^ i-h NroO r^ tJ- « On r^NO ^f 
O r^od on tj- « onoo' rA.vo" l o i o444cofoncoN ci c4 n 

IO M HH M HH 


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< 


O «oO\Tt qn^moo lOrfco Tf no ONCoONr^r^ONcoO OO 

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OOOOOOhhi-ih-ii-ihhi-iWMMMNCOCOCOCOCO'^- 

666666666666 6 6 6 6 6 6 6 6 6 6 6 


< 

H 

C 

U 


C/3 


Q »0 0> COOO N to On m tJ-nO 00 On O OnOO VO ^OVO O "^"^ 
O t^Tj-M ONt^Tt"-! ON NO COO NIOmOO »om OMONOO -^t 
O m fO lONO 00 O N rO^NO\0 M 'tiONCNO M rJ-iOt^ 

oqpoqoMHHMHHWMC^MNMNMcocpcococo 
66666666666666666666666 


O 

U 


6 a 


o h n f^Tj- vono r^oo ono h n nt lONO t^00 ONO h N 

HH h-t t-H H4I-IMHHI-II-II-CC4MM 






INDEX 



PAGE 

Acceleration of Gravity ... 74 

Active Pressure 234, 235 

Activity 45 

Air Gap Density 193 

Alternating Current Circuits, 261, 265 
Calculation of Single Phase, 276, 277 
Calculation of Three Phase . . 279 
Calculation of Two Phase . . 278 
Alternating Current Wiring, Ex- 
amples in 269, 270 

Formulas and Tables for . 272, 275 

Alternating Currents 230 

Ohm's Law of 236 

Original Problems in .... 251 
Alternators, E.M.F. of . . 150, 232 

Ampere, Absolute 6 

Definition of, by International 

Congress 7 

Practical 6 

Ampere-Turns 113 

Angle of Lag 236 

Angle of Lead .... 195, 196, 240 
Angle of Span of Poles .... 192 
Angular Velocity and Active Pres- 
sure 235, 236 

Anode 17 

Arc Lamps 66 

Arc Machines, Flux Density in . 193 

Arcs, Incandescent 66 

Armature Cross Section Relative 

to Field 193 

Armature Lamina? 192 

Magnetization 194 

Proportions 192 



PAGE 

Armatures, Calculation of Bipolar, 
Direct Current Machines, 

206, 207 

Of JH. P. Bipolar .... 222 

Of 10 K.W. Four-Pole . 215, 216 

Eddy Current Loss in ... . 198 

E.M.F. in, 158, 159, 160, 161, 162, 198 

Energy Lost in ... . 198, 202 

Examples in Calculation of . . 163 

Hysteresis Loss in 199 

I 2 R Loss in 191,208 

Mesh Connection of .... 160 

Monocyclic 161 

Peripheral Speeds of .... 192 

Polyphase 157 

Reaction in 195 

Reduction of Area by Lamina- 
tion 185 

Size of Wire for 191 

Star Connection of 161 

Surface Relative to Losses . . 200 

Temperature of ... . 191, 192 

Three-Phase 160 

Two-Phase 157 

Atomic Weight 98 

Average Pressure . . 160, 231, 232 

Back Turns 195, 196 

Compensation for . . . 197, 213 

Batteries 85 

Arrangement for Maxium Cur- 
rent 90 

For Required Efficiency . . 94 

Best Arrangement 89 



3OI 



302 



INDEX. 



PAGE 

Batteries, Best Arrangement for 

Required Current .... 91 
Batteries, Comparison of Parallel 

and Series Connection . . 88 
Comparison of Parallel Series 
with Parallel and Series 

Connections 89 

Connection for Combined Output, 85 

Consumption of Material in . . 102 

Diagram of Connection ... 89 

Original Problems in ... . 105 

Series and Multiple Connection, 89 

B-H Curves 229 

Bipolar, Direct Current Machine . 148 

Calculation of Armature . 206, 207 

Calculation of Fields .... 209 

Design of 206 

Brush Friction 199 

Calculation of Consequent Pole 
Fields 176 

Calculation of Fields . . . 171, 202 

Multipolar Type 180 

Original Problems in .... 185 
Calories, Kilogram and Gram 99, 100 

Relation to Ergs 50 

Relation to Joules 49 

Capacities in Series 242 

Capacity 237 

And Inductance . . . 241,242,249 
Inductance and Resistance 238, 245 
Quantity and E.M.F .... 13 

Relation to Circuit 238 

Resistance in Series with . . . 240 
Resistance in Parallel with . . 248 

Safe Carrying 53 

Unit of 10 

Cathode 17 

Cells, Arrangement for Maximum 

Current 90 

For Required Efficiency . . 94 

Best Arrangement of ... . 89 

For Required Current ... 91 

E.M.F. from Available Heat . 97 

Internal Drop 36 



PAGE 

Cells, Material Consumed in . . 102 
Measurement of E.M.F. of . . 36 
Cells, Multiple-Series Connection. 88 
Rules for Arrangement for Re- 
quired Current 92 

Storage, Charging 96 

Centimeter 3 

Chemical Action, E.M.F. from 

Available Heat of ... . 97 
Chemical and Electrochemical 

Constants 100 

Chemical Equivalent . . . . . 98 

Circuit, Magnetic no 

And Ohm's Law 123 

Circuits, Alternating Current . . 261 
Calculation of Long Distance, 

269, 292, 294 

Circular Mils 22 

Clark Cell 9 

Coefficient of Leakage . 113,182,202 

In Consequent Pole Type . . 176 

In Multipolar Type 180 

Coefficient of Self-induction, 235, 236, 
243, 244, 245 
Coefficient, Temperature of Mag- 
netism 137 

Coefficient, Temperature of Resis- 
tance .30 

Commercial Efficiency . . .54, 199 

Commutating Plane 195 

Compound Circuits 32 

Compounding . 196 

Condenser 237, 240 

Conductance and Conductivity . 30 

Conductance of Parallel Circuits . 33 

Conductors, Temperature of . . 51 
Connection of Cells for Combined 

Output 85 

Connection of Cells in Parallel . 86 

In Series 86 

Conservation of Energy, Law 

of 233 

Constant Current System ... 66 

Constant Potential System ... 67 

Continuous Current Machine . . 171 



INDEX. 



303 



PAGE 

Copper and Core Losses .... 262 
Copper, Electrochemical Equiva- 
lent of 17 

Copper, Melting Point of . . . 51 

Copper Sulphate 17 

Core Losses and Effect on Ratio 

of Transformation. . . . 262 

Coulomb 8 

Coulomb-Hour, Battery Material 

Consumed in 104 

Counter E.M.F. ... 58, 59, 234 
Cross Section of Fields and Arma- 
ture 193 

Cross Turns 195, 196 

Compensation for . . . 196, 213 
Current Intensity in Compound 

Circuits 34 

Current in the Three Wire System 78 

Magnetic Relations of ... . 15 

Relation of Heat to . . . . 48 

Unit of 6 

Currents, Alternating 230 

Eddy and Hysteresis . ... 139 

Eddy or Foucault 115 

Electrolytic Effects of ... . 16 

Curves, B-H 229 

Sine 230, 231 

Cycle 232 

Daniell Cell, Material Consumed 

in 103 

Data, Useful, in Designing Dyna- 
mos 190 

Definitions in Alternating Cur- 
rents 230 

In Electrical and Magnetic 

Qualities 1 

In Magnetism no 

Delta Connection of Armatures, 

32, 52 

Density of Magnetic Flux . no, 193 

Specific 26 

Design, Elements of Dynamo . . 190 

General Considerations . . . 190 

Illustrative Examples in ... 199 



PAGE 

Design of Bipolar, Direct Current 

Machine 206 

Of Four-Pole Dynamo . . . 215 
Of One- Fourth H.P. Bipolar 

Motor 199 

Diagrams and Wiring Formulae . 71 

For Wiring 76 

Of Cells in Multiple . . . 86, 87 
Multiple-Series .... 88, 89 

Series 153, 154 

Of Monocyclic Circuits . . . 285 
Of Monocyclic, Three Wire Sec- 
ondary 287 

Of Three Phase Circuits ... 281 
Of Three Phase, Four Wire Cir- 
cuits 283 

Of Two Phase Circuits ... 278 
Difference of Potential . . . 34, 66 
Direct Current Bipolar Machines, 148 
Direct Current Circuits, Calcula- 
tion of Three Wire . . . 276 
Direct Current Circuits, Calcula- 
tion of Two Wire .... 274 
Drop, Differences on Changing 

Load 75 

Distribution of, in Circuits . . 71 
In two Phase, Three Wire Cir- 
cuits 279 

In Wiring 76 

Formulae for . 197 

Reasons for Small 75 

Duty 54 

Dynamo, Calculation of Armature 

of a 5 K.W 207 

OfaioK.W 216 

Of Fields of a 10 K.W. ... 217 
Design of Bipolar, Direct Cur- 
rent 206 

Elements of Design 190 

Dynamos and Motors, E.M.F. of, 

148, 150, 153, 157, 158, 159, 160, 161 

I 2 R Loss in 191 

Output Relative to Weight . . 190 
Table of I 2 R Armature and Field 

Losses 192 



304 



INDEX. 



PAGE 

Dynamos and Motors, Tempera- 
ture of 191 

Dyne 2 

Eddy Currents 115 

And Hysteresis .... 139, 215 

In Armatures 198 

Effective Pressure .... 160, 232 
Efficiency, Arrangement of Cells 

for Required 94 

Calculation of Dynamo . 206, 215 
Commercial, of Dynamos . . . 199 
Electrical, of Dynamos . . 54, 199 
Full Load and Half Load ... 57 

Of Conversion 54 

Of Dynamos, Motors, and Trans- 
formers 57, 58 

Of Engines 57 

Of Motors, relation to Output . 59 
Of Transformation and Trans- 
mission 54 

Of Transformers 262 

Electrical Energy 45 

And Heat 47 

And Mechanical Energy . . 45,46 
Original Problems in ... . 59 

Relation to Heat 99 

Electrical Quantities, Original 

Problems in 18 

Electrical Units 6 

Electrochemical and Chemical 

Constants 100 

Electrochemical Eq ui valents , 8 , 1 7 , 1 8 , 99 
Electrolytic Effects of Currents . 16 
Electromagnetic Units .... 6 
Electromotive Force, Unit of . . 8, 9 

Electrostatic Units 6 

Elements of Dynamo Design . . 190 
Illustrative Examples in . . . 199 

Useful Data in 190 

E.M.F. and Heat of Cells ... 100 

And Magnetic Flux 148 

And Rate of Change of Lines of 

Force 234, 242 

Average 231 

Counter 5 8 > 59> 2 34 



PAGE 

E.M.F., Frequency of 232 

Generated in Armatures . . . 198 
In Three Wire System .... 78 

Maximum 232, 234 

Measurement of Drop of, in 

Cells 36 

Measurement of, in Cells . . 36 
Of Alternators ...... 150 

Of Bipolar Direct Current Ma- 
chines 148 

Of Cells from Available Heat . 97 

Of Clark Cell 9 

Of Dynamos and Motors . . . 148 
Of Dynamos, Original Problems 

in 166 

Of Monocyclic Machines . 162, 163 
Of Multipolar Machines . . . 153 
Of Self-induction 233, 234, 243, 244 
Of Three Phase Machines 160, 161 
Of Two Phase Machines ... 158 

Period of 232 

Quantity and Capacity .... 13 

Relation to Angle 230 

Energy Due to Magnetization . . 11 1 

Erg 2,45 

Ergs, Relation to Calories ... 50 

Farad 10 

Field, Calculation of . . . 171, 202 
Calculation of Bipolar, Direct 

Current .... 202, 209, 217 
Cross Section Relative to Arma- 
ture 193 

Density of Flux in 193 

Drop 197 

Intensity 4, " r 

I 2 R Loss in 191, 192 

Magnetic 4 

Of One-Fourth H.P. Motor . . 222 

Size of Wire for 191 

Watts Lost in ....... 198 

Winding, Shunt 196 

Flux densities 193 

Lines, Length of 205 

Magnetic ...... no, 120 



INDEX. 



305 



PAGE 

Foot-Pound 45 

Force, Law of Magnetic .... 116 

• Lines of 4 

Magnetizing 113 

Magnetomotive 112 

Formulae and Diagrams for Wir- 
ing 71 

And Tables for Alternating 

Current Wiring 272 

For Efficiency of Cells ... 95 

Simplified for Wiring .... 74 

Useful in Designing .... 197 

Foucault Currents 115 

Four-Pole Dynamo, Calculation 

of Fields of 217 

Design of 215 

Frequency of Alternating E.M.F. 

232, 233 

Friction of Brushes 199 

Fuse Wire, Size of 52 

Gausses no, m 

General Laws of Resistance . . 22 

Geometric Sum 246 

Gilberts 112 

Gram-Degree 49 

Gravity, Acceleration of ... . 47 

H 4 . 4, in, 113 

Heat and Electrical Energy . . 47, 99 
E.M.F. of Cells from Available, 

97, 100 
Rate of Radiation of ... . 51 
Relation to Current .... 48 

To E.M.F 48 

To Mechanical Work ... 49 

To Resistance 48 

Unit of 49 

Heats of Formation and Separa- 
tion 102 

Henry 11,237 

Horse Power J i,45 

Hydrogen, Electrochemical Equiv- 
alent of 18 

Hysteresis and Eddy Currents , . 139 



PAGE 

Hysteresis Loss in Armatures . . 199 
Magnetic u 4 

Illustrative Examples in Dyna- 
mo Design 199 

Impedance 236, 263 

Factor 268 

Impressed E.M.F. and Inductance 268 

Pressure 234, 236 

Incandescent Arcs 66 

Inductance, Capacity and Resist- 
ance 238, 245 

And Capacity in Parallel . . . 249 
In Series 241 

And Resistance in Parallel 245, 246 
In Series 238, 239 

Factor of Line .... 267, 268 

Unit of 237 

Inductances in Parallel . . 246, 249 

Series 239 

Induction, Coefficient of ... . 235 

Magnetic 5, no 

Mutual 233 

Self 233, 234 

Unit of n 

Inductive Pressure .... 234, 263 

Resistance 235, 236 

Intensity of Current in Com- 
pound Circuits 34 

Of Magnetic Field in 

Of Magnetic Flux no 

And Strength of Field . . . 120 

Of Magnetization 5 

International Ampere 7 

Ohm 10 

Volt 9 

Ions 101 

Heat of 102 

I 2 R Loss in Armatures . . 192, 208 

In Dynamos 192 

Joule 49 

Joule's Equivalent . . . . 49, 99 

Joules J 1, 46 

Relation to Calories , , , . 49 



3o6 



INDEX. 



PAGE 

Kilogram 3 

Kilogramme des Archives ... 3 

Kilogram-Meter 45 

Kilowatt 45 

I* 243, 245 

Lag, Angle of 236 

Caused by Line Inductance . . 267 
Of E.M.F. of Self-induction . 234 

Of Magnetism 114 

Laminae of Armature . . 185,192,201 

Lamp-Feet 73 

Lamps, Arc and Incandescent . . 66 

Poor Economy of 50 Volt . 76, 77 

Law of Conservation of Energy . 233 

Of Magnetic Force 116 

Of Resistance ....... 22 

Ohm's 16 

Lead, Angle of ... . 195, 196, 240 
Leads and Lamp Circuits, Calcula- 
tion of 67, 76 

Size of 66 

Leakage, Coefficient of . . 113,202 

Magnetic 113 

Length and Area, Relation to Re- 
sistance 22 

Lenz's Law 233 

Leyden Jar 11 

Lifting Power of Magnets . . . 132 
Thompson's Rule for . . 134, 135 
Light and Power, Wiring for . . 66 
Lights and Motors, Wiring Mono- 
cyclic Circuits for ... . 287 

Lines of Force 4 

Lines of Force and Current . . . 242 

And E.M.F 243 

Formula for . 243 

Rate of Change of 234 

Long Distance, Calculation of 

Circuits for . . . 269, 270, 293 

Loss in Armature 192, 202 

I' 2 R in Armature and Field . . 202 
Plant 54 

Machines, Bipolar Direct Cur- 
rent 148, 171 



PAGE 

Machines, Consequent Pole Type, 176 

Magnetic Circuit and Ohm's Law, 123 

Magnetic Field 4 

Intensity of m 

Magnetic Force, Law of . . . . n6 
Measurement of . . . 115,118,119 

Magnetic Flux .110 

Density of no 

Intensity of, and M.M.F. . . 114 
Intensity of Magnetism and 

Strength of Field . . . 120 

Lag of 114 

Leakage of 113 

Magnetic Hysteresis . . . . . 114 

Magnetic Induction 5 

Magnetic Moment 5 

Magnetic Properties, Table of, 229, 298 
Magnetic Quantities, Original 

Problems in J40 

Relation of 116 

Magnetic Relations of Current . 15 

Magnetic Reluctance 112 

Magnetic Units 4 

Magnetism and Temperature . . 137 

Remanent or Residual ....114 

Units and Definitions in . . . no 

Magnetization of Armature . . . 194 

Energy Due to in 

Intensity of ....... 5 

Magnetizing Force 113 

Magnetomotive Force 112 

Magnets, Lifting Power of . . . 132 

Thompson's Rule for . 134, 135 

Saturated 114 

Manganin 28 

Mass 3 

Material Consumed in Cells . . 102 
Maximum Current, Arrangement 

of Cells for 89, 90 

Maximum E.M.F. in Two Phase, 

Three Wire Circuits . . . 280 

Maxwells of Flux . ...... no 

Maxwell's Formula in 

Mechanical Work, Relation to 

Heat 49 

Megohm 10 



INDEX. 



307 



PAGE 

Mesh Connection of Armatures . 160 

Formula for E.M.F. of ... 160 

Meter 2 

Meter des Archives 2 

Microcoulomb 8 

Microfarad 10 

Microhm 10 

Mil 22 

Mil-Foot 22 

Millimeter 3 

Minimum E.M.F. in Two Phase, 

Three Wire Circuits . . . 280 

Monocyclic Armatures . . . . 161 

E.M.F. of 162 

Monocyclic Circuits 284 

Connection of Transformers in . 162 

Three Wire Secondary .... 287 

Winding of Transformers for . 163 
Motor, Calculation of Armature of 

\ H.P 200, 222 

Of Fields 202 

Design of \ H.P 199 

Of \ H.P 222 

Pump, Power for 65 

Motors, E.M.F. of 148 

Output and Efficiency of . . 58, 59 

Power Factor of 268 

Multiple Connection of Cells . . 86 
Multiple-Series Connection of 

Cells 88 

Compared with Multiple and 

Series 89 

Multiple Winding . " . . . 153, 154 

Multipolar Dynamo, E.M.F. of . 154 

Multipolar Type 180 

Coefficient of Leakage of . . . 180 

Mutual Induction 233 

Neutral Wire 78, 79 

Non-inductive Loads and Line 

Induction 265 

Number of Turns and E.M.F . . 242 

Oersted . 112 

Ohm 10 



PAGE 

Ohm, International 10 

Ohmic Resistance 236 

Ohm's Law 12 

And the Magnetic Circuit . . 123 
Ohm's Law Applied to the Electric 

Cell 85 

Of Alternating Current Circuits, 236 

Ordinates 230 

Original Problems in Alternating 

Currents 251 

In Alternating Current Distribu- 
tion 290 

In Batteries 105 

In Calculation of Fields . . . 185 
In Electrical Energy .... 59 
In E.M.F. of Dynamos . . . .166 
In General Laws of Resistance . 36 
In Relation of Electrical Quan- 
tities . 18 

In Relation of Magnetic Quan- 
tities 140 

In Wiring 80 

Output of Dynamos, Relation to 

Weight 190 

Over-Compounding 197 

Calculation of 214 

Oxygen, Chemical Equivalent of . 98 

Electrochemical Equivalent of . 99 

Parallel Circuits 32 

Conductance of 33 

Resistance of 33, 245 

Parallel Connection of Cells . . 87 

Parallel System 67 

Period of an Alternating E.M.F., 232 
Peripheral Speeds of Armatures . 192 

Permeability 5, 112, 243 

Permeameter Method 121 

Permeance 5 

Phase 235 

Plant Efficiency 54 

Pole, Strength of 4, 116 

Unit 4, 115 

Polyphase Armatures, 157, 159, 160, 161 
Calculation of . . . , 163,164,166 



3o8 



INDEX. 



PAGE 

Potential, Difference of ... . 34 

Drop of, and Resistance ... 34 

And Size of Leads .... 66 

Power and Work 14 

Power Factor of Line 267 

With Motors - . 268 

Power of Magnets 132 

Required for Car . . . 57,58,65 

For Pump . 65 

Unit of 11 

Pressure, Active . . . . 234, 235 

Average 150,231 

Due to Capacity 237 

Effective 150, 232 

Impressed 234, 237 

Self-Inductive .... 234, 243 

Problems in Alternating Currents, 251 
In Alternating Current Distribu- 
tion 290 

In Batteries 105 

In Electrical Energy .... 59 
In Elements of Dynamo Design, 199 
In E.M.F. of Dynamos ... 166 
In Field Calculation . . . . 185 
In General Laws of Resistance, 36 
In Relation of Electrical Quan- 
tities 18 

In Relation of Magnetic Quan- 
tities 140 

In Wiring 80 

Proportions of Armatures . . . 192 

Quantities, Relation of . . . 12, 116 

Quantity and Capacity .... 13 

Unit of 8 

Radiation of Heat, Rate of . . 51 
Rate of Change of Lines of Force, 

234, 243 
Rate of Work, Maximum for 

Motor 59 

Ratio of Transformation .... 261 
Effect of Core Losses on . . . 262 

Reactance 236, 240 

Reaction of Armatures .... 195 



PAGE 

Relation of Electric Quantities . 12 

Original Problems in ... 18 

Of Magnetic Quantities . . . 116 

Original Problems in ... 140 

Reluctance 5 

Magnetic 112 

Reluctivity 112 

Constants of 126 

Formulae for 197 

Remanent or Residual Magnetism, 114 
Resistance and Capacity in Par- 
allel 248 

In Series 240 

And Drop of Potential . . . 34, 72 
And Inductance in parallel and 

series 239, 245 

Due to Capacity 237 

General Laws of 22 

Increase with Rise of Temper- 
ature 28, 183, 200 

Inductance and Capacity . . 238, 245 

Inductive 235, 236^ 

Of Lamps and Wire 72 

Of Parallel Circuits . .33,85,245 
Original Problems in General 

Laws of 36 

Practical Standards of ... . 10 

Relation to Heat 48 

To Length and Area ... 22 

To Temperature 27 

To Weight of Wire .... 26 

Table of Specific 25 

Temperature Coefficients of . 28, 30 

Unit of 10 

Ring Armature, Calculation of, for 

4-Pole Dynamo 215 

Rules for Arrangement of Cells . 92 , 93 

Safe Carrying Capacity ... 53 
Saturation of Magnets .... 114 

Second 3 

Self-Induction, Coefficient of . . 

235,236,243, 244 

Effect on Line 266 

E.M.F. of 233,234,244 



INDEX. 



309 



PAGE 

Self-induction, Measurement of 

Coefficient of 245 

Resistance Due to 235 

Series Connection of Cells ... 85 

Series Field Drop 197 

Watts Lost in 19 8 

Series-Multiple Connection of 

Cells 88 

Series Winding of Armatures . .154 
Shunt Field, Watts Lost in . . . 198 

Winding of *9 6 

Shunts 32 

Silver, Electrochemical Equiva- 
lent of 17 

Silver Nitrate 17 

Sine Curve ...... .230,231 

Single Phase Circuits, Calculation 

of 276, 277 

Solenoid 16 

Span of Poles 192 

Specific Density 26 

Specific Gravity ....... 25 

Specific Heat 25, 50 

Specific Reluctance 112 

Specific Resistance . . . 24, 25, 71, 72 
Speeds, Peripheral of Armatures . 192 

Square Mils 22 

Star Connection of Armatures . . 161 
Formula for E.M.F. of . . . 161 

Station Efficiency 54 

Steinmetz's Formula . . . . 114, 115 
Street Car, Power Required for, 57, 65 
Strength of Field, Magnetic Flux 

and Intensity of Magnetism 120 

Strength of Pole 4, 116 

Storage Cells, Charging .... 96 

Table of Alternating Current 

Wiring Constants . .273,275 
Of Chemical and Electrochemi- 
cal Constants 100 

Of Heats of Chemical Action . 102 

Of Hysteresis Losses .... 199 

Of I 2 R Losses 192 

Of Magnetics Properties . . . 298 



PAGE 

Table of Resistance, Inductance, 

and Impedance .... 265 

Reference to 263 

Of Specific Resistance, Specific 

Gravity, and Specific Heat . 25 
Of Trigonometric Functions . . 299 
Of Wiring Constants, 70, 265, 273, 275 

Teazer 162, 287 

Temperature and Magnetism . . 137 
Coefficient of Magnetism . . . 138 

Temperature Coefficient of Resist- 
ance 28, 30 

Effect on Field Resistance . . 183 

Of Conductors 51 

Relation to Resistance .... 27 

Thermal Equivalents 99 

Three Phase Armatures .... 160 
Formulae for E.M.F. and Cur- 
rent in 160 

System, Four Wire Distribution 282 
Three Wire Transmission . . 281 

Three Wire Circuits, Calculation of 276 
Size of Wire for 78, 79 

Three Wire System jj 

Connection of Lamps on . . . 78 

Current in 78 

E.M.F. of 78 

Time Constant 238, 246 

Transformation and Transmission, 

Efficiency of 54 

Ratio of 261, 262 

Transformers, Copper Loss in . . 262 

Core Loss in 262 

Efficiency of 262 

On Monocyclic Circuits, 162, 163, 287 

Turns, Number of, and E.M.F., 

148, 242 

Two Phase Armatures .... 157 
Formula for E.M.F. and Cur- 
rent in 158, 159 

Two Phase Four Wire Circuits . 278 
Three Wire Distribution . 279, 280 

Unit of Capacity 10 

Of Current 6 



3io 



INDEX. 



PAGE 

Unit, E.M.F. ..*.... 8 

Of Heat 49 

Of Induction n 

Of Magnetism no 

Of Power n 

Of Quantity 8 

Of Resistance io 

Of Work ii 

Unit Pole 4, 116 

Units, Absolute 2 

Units, Basis of Fundamental . . 2 

Electrical '. . 6 

Electromagnetic 6 

Electrostatic 6 

Fundamental and Derived . . 1 

Magnetic 4 

Valence 98 

Volt 9 

Volt-Coulombs 46 

Volt, International 9 

Volts, Drop 7i)7S 

Formula for 197 

Water-Gram-Degree .... 49 

Watt n,45 

Watt-Hour, Material Consumed in 

Cell per 103 

Watts in One Horse-Power ... 47 



PAGE 

Watts, Relation to Calories ... 50 

Wave Winding 154 

Weight 3 

Weight, Relation to Resistance . 26 

Windage 199 

Winding, Multiple .... 153, 1*54 

Series 154 

Wiring, Alternating Current, 

Monocyclic .... 284, 287 
Constants, Table of 

70, 265, 273, 275 

Diagrams 76 

Direct Current, Three Wire . . 276 

Two Wire 274 

For Light and Power .... 66 

Formulae and Diagrams ... 71 

Formula Simplified 74 

Original Problems in, Alternat- 
ing Current 290 

Direct Current 80 

Rules for Correct 68 

Single Phase 276, 277 

Three Phase, Four Wire ... 282 

Three Wire 281 

Two Phase, Four Wire ... 278 

Three Wire 279 

Work and Power 14 

Unit of « 



LIST OF WORKS 

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CROCKER, F. B., and S. S. WHEELER. The Practical Management of 
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5 



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*** A General Catalogue— 80 pages—of Works in all branches of 
Electrical Science furnished gratis on application, 






1 



APR 4 1913 



